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arXiv:1107.1130v1 [math.NT] 6 Jul 2011

Dismal Arithmetic

David Applegate,

AT&T Shannon Labs,

180 Park Ave., Florham Park, NJ 07932-0971, USA

david@research.att.com

Marc LeBrun,

Fixpoint Inc.,

448 Ignacio Blvd. #239, Novato, CA 94949, USA

mlb@well.com

N. J. A. Sloane,(a)

AT&T Shannon Labs,

180 Park Ave., Florham Park, NJ 07932-0971, USA

njas@research.att.com

(a)To whom correspondence should be addressed.

To the memory of Martin Gardner (October 21, 1914 – May 22, 2010).

July 5, 2011

Abstract

Dismal arithmetic is just like the arithmetic you learned in school, only simpler: there there

are no carries, when you add digits you just take the largest, and when you multiply digits

you take the smallest. This paper studies basic number theory in this world, including

analogues of the primes, number of divisors, sum of divisors, and the partition function.

1 Introduction

To remedy the dismal state of arithmetic skills possessed by today’s children, we propose a

“dismal arithmetic” that will be easier to learn than the usual version. It is easier because

there are no carry digits and there is no need to add or multiply digits, or to do anything

harder than comparing. In dismal arithmetic, for each pair of digits,

to Add, take the lArger, but

to Multiply, take the sMaller.

That’s it! For example: 2 5 = 5, 2 5 = 2.

Addition or multiplication of larger numbers uses the same rules, always with the proviso

that there are no carries. For example, the dismal sum of 169 and 248 is 269 and their dismal

product is 12468 (Figure 1).

1

169

248

269

Fig. 1(a) Dismal addition.

169

248

168

1 4 4

1 2 2

12468

Fig. 1(b) Dismal multiplication.

One might expect that nothing interesting could arise from such simple rules. However,

developing the dismal analogue of ordinary elementary number theory will lead us to some

surprisingly diﬃcult questions.

Here are a few dismal analogues of standard sequences. The “even” numbers, 2 n, are

0,1,2,2,2,2,2,2,2,2,10,11,12,12,12,12,12,12,12,12,20,21,22,22,··· (1)

(entry A171818 in [17]). Note that n n (which is simply n) is a diﬀerent sequence. For

another, less obvious, analogue of the even numbers, see (13) in §3. The squares, n n, are

0,1,2,3,4,5,6,7,8,9,100,111,112,113,114,115,116,117,118,119,200,··· (2)

(A087019),1the dismal triangular numbers, 0 1 2 . . . n, are

0,1,2,3,4,5,6,7,8,9,19,19,19,19,19,19,19,19,19,19,29,29,29,29,29, . . . (3)

(A087052), and the dismal factorials, 1 2 ··· n,n≥1, are

1,1,1,1,1,1,1,1,1,10,110,1110,11110,111110,1111110,11111110,111111110,

1111111110,11111111110,111111111100,1111111111100,11111111111100, . . . (4)

(A189788).

A formal deﬁnition of dismal arithmetic is given in §2, valid for any base b, not just

base 10, and it shown there that the commutative, associative, and distributive laws hold

(Theorem 1). In that section we also introduce the notion of a “digit map,” in order to

study how changing individual digits in a dismal calculation aﬀects the answer (Theorem 3,

Corollary 4).

The dismal primes are the subject of §3. A necessary condition for a number to be a

prime is that it contain a digit equal to b−1. The data suggest that if kis large, almost all

numbers of length kcontaining b−1 as a digit and not ending with zero are prime, and so

the number of primes of length kappears to approach (b−1)2bk−2as k→ ∞ (Conjecture 1).

In any case, any number with a digit equal to b−1 is a product of primes (Theorem 7), and

every number can be written as rtimes a product of primes, for some r∈ {0,1,...,b−1}

1See also the sums of two squares, A171120. The numbers 10,11,...,99 are not the sum of any number

of squares, so there is no dismal analogue of the four-squares theorem.

2

(Corollary 8). These factorizations are in general not unique. There is a useful process using

digit maps for “promoting” a prime from a lower base to a higher base, which enables us to

replace the list of all primes by a shorter list of prime “templates” (Table 3).

Dismal squares are brieﬂy discussed in §4.

In §5we investigate the diﬀerent ways to order the dismal numbers, and in particular

the partially ordered set deﬁned by the divisibility relation (see Table 1). We will see that

greatest common divisors and least common multiples need not exist, so this poset fails to

be a lattice. On the other hand, we do have the notion of “relatively prime” and we can

deﬁne an analogue of the Euler totient function.

In §6we study the number-of-divisors function db(n), and investigate which numbers

have the most divisors. It appears that in any base b≥3, the number n= (bk−1)/(b−1) =

111 ...1|bhas more divisors than any other number of length k. The binary case is slightly

diﬀerent. Here it appears that among all k-digit numbers n, the maximal value of d2(n)

occurs at n= 2k−2 = 111 ...10|2, and this is the unique maximum for n6= 2,4. Among

all odd k-digit numbers n,d2(n) has a unique maximum at n= 2k−1 = 111 ...111|2,

and if k≥3 and k6= 5, the next largest value occurs at n= 2k−3 = 111 ...101|2, its

reversal 2k−2k−2−1 = 101 ...111|2, and possibly other values of n(see Conjectures 2-4).

Although we cannot prove these conjectures, we are able to determine the exact values of

db(111 ...111|b) and d2(111 ...101|2) (Theorem 13, which extends earlier work of Richard

Schroeppel and the second author, and Theorem 14).

The sequence of the number of divisors of 11 ...11|2(with k1’s) turns out to arise in a

variety of diﬀerent problems, involving compositions, trees, polyominoes, Dyck paths, etc.—

see Remark (iii) following Theorem 12. The initial terms can be seen in Table 8. This

sequence appears in two entries in [17], A007059 and A079500, and is the subject of a survey

article by Frosini and Rinaldi [6]. The asymptotic behavior of this sequence was determined

by Kemp [10] and by Knopfmacher and Robbins [12], the latter using the method of Mellin

transforms—see (23). This is an example of an asymptotic expansion where the leading

term has an oscillating component which, though small, does not go to zero. It is amusing

to note that one of the ﬁrst problems in which the asymptotic behavior was shown to involve

a nonvanishing oscillating term was the analysis of the average number of carries when two

k-digit numbers are added (Knuth [13], answering a question of von Neumann; see also

Pippenger [18]). Here we see a similar phenomenon when there are no carries. In studying

Conjectures 3and 4, we observed that the numbers of divisors for the runners-up, 2k−3 and

2k−2k−2−1, appeared to be converging to one-ﬁfth of the number of divisors of 11 ...11|2.

This is proved in Theorem 19. Our proof is modeled on Knopfmacher and Robbins’s proof

[12] of (23), and we present the proof in such a way that it yields both results simultaneously.

The sum-of-divisors function σb(n) is the subject of §7. There are analogues of the

perfect numbers, although they seem not to be as interesting as in the classical case. Section

8discusses the dismal analogue of the partition function. Theorems 22 and 23 give explicit

formulas for the number of partitions of ninto distinct parts.

This is the second of a series of articles dealing with various kinds of carryless arithmetic,

and contains a report of our investigations into dismal arithmetic carried out during the

period 2000–2011. This work had its origin in a study by the second author into the results

of performing binary arithmetic calculations with the usual addition and multiplication of

3

binary digits replaced by other operations. If addition and multiplication are replaced by the

logical operations OR and AND, respectively, we get base 2 dismal arithmetic. (If instead

we use XOR and AND, the results are very diﬀerent, the squares for example now forming

the Moser-de Bruijn sequence A000695.) Generalizing from base 2 to base 10 and then to

an arbitrary base led to the present work.

In the ﬁrst article in the series, [1], addition and multiplication were carried out “mod

10”, with no carries. A planned third part will discuss even more exotic arithmetics.

Although dismal arithmetic superﬁcially resembles “tropical mathematics” [20], where

addition and multiplication are deﬁned by x⊕y:= min{x, y},x⊙y:= x+y, there is no

real connection, since tropical mathematics is deﬁned over R∪ {∞}, uses carries, and is not

base-dependent.

Notation. The base will be denoted by band the largest digit in a base bexpansion by

β:= b−1. We write n=nk−1nk−2...n1n0|bto denote the base brepresentation of the

number Pk−1

i=0 nibi, and we deﬁne lenb(n) := k. The components niwill be called the digits

of n, even if b6= 10. In the examples in this paper bwill be at most 10, so the notation

nk−1. . . n1n0|b(without commas) is unambiguous. The symbols band bdenote dismal

addition and multiplication, and we omit the base bif it is clear from the context. All

non-bold operators (+, ×,<, etc.) refer to ordinary arithmetic operations, as do unqualiﬁed

terms like “smallest,” “largest,” etc. We usually omit ordinary multiplication signs, but

never dismal multiplication signs. We say that pdivides nin base b(written p≺bn) if

pbq=nfor some q, and that p=pk−1. . . p1p0|bis dominated by n=nk−1. . . n1n0|b

(written p≪bn) if pi≤nifor all i. The symbol “|b” always marks the end of a base b

expansion of a number, and is never used for “divides in base b.”

2 Basic deﬁnitions and properties

We began, as we all did, in base 10, but from now on we will allow the base bto be an

arbitrary integer ≥2.

Let Adenote the set of base b“digits” {0,1,2,...,b−1}, equipped with the two binary

operations

mbn:= max{m, n}, m bn:= min{m, n},for m, n ∈ A .(5)

Adismal number is an element of the semiring A[X] of polynomials Pk−1

i=0 niXi,ni∈ A. If

M[X] := Pk−1

i=0 miXiand N[X] := Pl−1

i=0 niXiare dismal numbers then their dismal sum is

formed by taking the dismal sum of corresponding pairs of digits, analogously to ordinary

addition of polynomials:

M[X]bN[X] :=

max{k,l}−1

X

i=0

piXi,(6)

where pi:= mibni, and their dismal product is similarly formed by using dismal arithmetic

to convolve the digits, analogously to ordinary multiplication of polynomials:

M[X]bN[X] :=

k+l−2

X

i=0

qiXi,(7)

4

where

q0:= m0bn0,

q1:= (m0bn1)b(m1bn0),

q2:= (m0bn2)b(m1bn1)b(m2bn0),

. . . .

We will identify a dismal number N(X) = Pk−1

i=0 niXiwith the integer nwhose base b

expansion is n=Pk−1

i=0 nibi(nis obtained by evaluating the polynomial N(X) at X=b),

and we deﬁne lenb(n) := k. The rules (5)-(7) then translate into the rules for dismal addition

and multiplication stated in §1: there are no carries, and digits are combined according to

the rules in (5). The b-ary number Pk−1

i=0 nibiwill also be written as nk−1nk−2...n1n0|b.

Dismal numbers are, by deﬁnition, always identiﬁed with nonnegative integers. Note that

lenb(mbn) = max{lenb(m),lenb(n)}and lenb(mbn) = lenb(m) + lenb(n)−1.

Theorem 1. The dismal operations band bon A[X]satisfy the commutative and asso-

ciative laws, and bdistributes over b.

Proof. (Sketch.) Each law requires us to show the identity of two polynomials, and so reduces

to showing that certain identities hold for the coeﬃcients of each individual degree in the

two polynomials. These identities are assertions about min and max in the set A, which hold

since (A,≤) is a totally ordered set, and (A,min,max) is a distributive lattice (cf. [9]).

If Rdenotes the operation of reversing the order of digits and mand nhave the same

length, then R(mbn) = R(m)bR(n) and R(mbn) = R(m)bR(n).

Individual digits in a dismal sum or product can often be varied without aﬀecting the re-

sult, so dismal subtraction and division will not be deﬁned. Example: 16 10 75 = 26 10 75 =

76, 16 10 75 = 16 10 85 = 165. This is why dismal numbers form only a semiring. On the

other hand, this semiring does have a multiplicative identity (see the next section), and there

are no zero divisors.

In certain situations we can give a more precise statement about how changing digits in

a dismal sum or product aﬀects the result. We begin with a lemma about ordinary functions

of real variables.

Lemma 2. Let fbe a single-valued function of real variables x1,...,xk,k≥2, formed by

repeatedly composing the functions (x, y)7→ min{x, y}and (x, y)7→ max{x, y}. If gis a

nondecreasing function of x, meaning that

x≤y⇒g(x)≤g(y),(8)

then

f(g(x1),...,g(xk)) = g(f(x1,...,xk)) ,(9)

for all real x1,...,xk.

We omit the easy inductive proof.

We deﬁne a base bdigit map to be a nondecreasing function gmapping {0,1,...,b−1}into

itself. The map gneed not be one-to-one or onto. If gis a digit map and n=nk−1. . . n1n0|b

then we set g(n) := g(nk−1). . . g(n1)g(n0)|b.

5

Theorem 3. If mand nare dismal numbers and gis a base bdigit map, then

g(mbn) = g(m)bg(n),

g(mbn) = g(m)bg(n).(10)

Proof. This follows from Lemma 2, since the individual digits of mbnand mbnare

functions of the digits of mand nof the type considered in that lemma.

Corollary 4. If p=mbnthen pcan also be written as m′bn′, where m′and n′use only

digits that are digits of p.

Proof. (Sketch.) Arrange all distinct digits occurring in p,m, and nin increasing order.

Then construct a digit map gby increasing or decreasing the digits in mand nthat are not

in puntil they coincide with digits of p, leaving the digits of pﬁxed.

For example, consider the product 165 = 16 10 85 mentioned above. The digits involved

are 1,5,6,8, and the digit map described in the proof ﬁxes 1,5,and 6 and maps 8 to 6. The

resulting factorization is 165 = 16 10 65. (The additive analogue of Corollary 4is true, but

trivial.)

We will see other applications of Theorem 3in the next section.

Note that when we are computing the base bdismal sum or product of two numbers p

and q, once we have expressed pand qin base b, the value of bplays no further role in the

calculation. Of course we need to know bwhen we convert the result back to an integer, but

otherwise bis not used. So we have:

Lemma 5. If the largest digit that is mentioned in a base bdismal sum or product is d

(where 0≤d≤b−1), then the same calculation is valid in any base that exceeds d.

For example, here is the calculation of the base 2 dismal product of 13 = 1101|2and

5 = 101|2:

1101

2101

1101

1 1 0 1

111101

This tells us that 13 25 = 61, but the same tableau can be read in base 3, giving 37 310 =

361, or in base 10, giving 1101 10 101 = 111101.

Recall that we say that pdivides nin base b(written p≺bn) if pbq=nfor some q.

Since lenb(p)≤lenb(n), nonzero numbers have only ﬁnitely many divisors. We also say that

p:= pk−1. . . p1p0|bis dominated by n:= nk−1. . . n1n0|b(written p≪bn) if pi≤nifor all i.

Then p≪bnif and only if pbn=n. Another consequence of Lemma 2is:

Lemma 6. If p≪bmand q≪bnthen pbq≪bmbnand pbq≪bmbn.

Finally, we remark without giving any details that, in any base, the sets of numbers with

digits in nondecreasing order, or in nonincreasing order (see A009994 and A009996 for base

10) are closed under dismal addition and multiplication.

6

3 Dismal primes

In dismal arithmetic in base b, for bases b > 2, the multiplicative identity is no longer 1 (for

example, 1 10 23 = 11, not 23). In fact, it follows from the deﬁnition of multiplication that

the multiplicative identity is the largest single-digit base bnumber, β:= b−1. For base

b= 10 we have β= 9, and indeed the reader will easily check that 9 10 n=nfor all n. An

empty dismal product is deﬁned to be β, by convention.

If pbq=β, then p=q=β, so βis the only unit. We therefore deﬁne a prime in base b

dismal arithmetic to be a number, diﬀerent from β, whose only factorization is βtimes itself.

If pis prime, then at least one digit of pmust equal β(for if the largest digit were r < β,

then p=rbp, and rwould be a divisor of p). The base bexpansions of the ﬁrst few primes

are 1β(this is the smallest prime), 2β, 3β . . . β −1β,β0, β1, . . . , ββ, 10β,.... In base 10, the

primes are

19,29,39,49,59,69,79,89,90,91,92,93,94,95,96,97,98,99,109,209,219,

309,319,329,409,419,429,439,509,519,529,539,549,609,619,629,639, . . . (11)

(A087097). Notice that the presence of a digit equal to βis a necessary but not suﬃcient

condition for a number to be a prime: 11β|b= 1β|bb1β|bis not prime (see A087984 for

these exceptions in the case b= 10). In base 2, the primes (written in base 2) are

10,11,101,1001,1011,1101,10001,10011,10111,11001,11101,100001, . . . (12)

(A171000). In view of the interpretation of base 2 dismal arithmetic in terms of Boolean

operations mentioned in §1, the corresponding polynomials

X, X + 1, X2+ 1, X3+ 1, X3+X+ 1, X 3+X2+ 1, X4+ 1, X4+X+ 1, . . . ,

together with 1, might be called the OR-irreducible Boolean polynomials. Their decimal

equivalents,

1,2,3,5,9,11,13,17,19,23,25,29,33,35,37, . . .

form sequence A067139 in [17], contributed by Jens Voß in 2002.

All numbers of the form 100 ...00β|b(with zero or more internal zeros) are base bprimes,

since there is no way that pbqcan have the form 100 ...00β|bunless por qis a single-digit

number. So there are certainly inﬁnitely many primes in any base.

Since 1β|bis the smallest prime, the numbers 1β|bbnare another analogue of the even

numbers. Whereas the ﬁrst version of the even numbers, given in (1) for base 10, was simply

“replace all digits in nthat are bigger than 2 with 2’s,” this version is more interesting. In

base 10 we get

0,11,12,13,14,15,16,17,18,19,110,111,112,113,114,115,116,117,118, . . . (13)

(A162672, which contains repetitions and is not monotonic).

If b > 2, there are numbers which cannot be written as a product of primes (e.g., 1).

Theorem 7. Any base bnumber with a digit equal to βis a (possibly empty) dismal product

of dismal primes.

7

Proof. The number βitself is the empty product of primes. Every two-digit number with β

as a digit is already a prime. If there are more than two digits, either the number is a prime,

or it factorizes into the product of two numbers, both of which must have βas a digit. The

result follows by induction.

Corollary 8. Every base bnumber can be written as rtimes a dismal product of dismal

primes, for some r∈ {0,1,2,...,b−1}.

Proof. Let rbe the largest digit of n. If r=βthe result follows from the theorem. Otherwise,

let mbe obtained by changing all occurrences of rin the b-ary expansion of nto β’s, so that

n=rbm, and apply the theorem to m.

Even when it exists, the factorization into a dismal product of dismal primes is in gen-

eral not unique. In base 10, for example, the list of numbers with at least two diﬀerent

factorizations into a product of primes is

1119,1129,1139,1149,1159,1169,1179,1189,1191,1192,1193,1194,1195, . . . , (14)

where for instance 1119 = 19 10 19 10 19 = 19 10 109 (A171004).

kbase 2 base 10 kbase 2 kbase 2 kbase 2

1 0 0 11 323 21 442313 31 510471015

2 2 18 12 682 22 902921 32 1027067090

3 1 81 13 1424 23 1833029 33 2065390101

4 3 1539 14 2902 24 3719745 34 4151081457

5 5 20457 15 5956 25 7548521 35 8336751732

6 9 242217 16 12368 26 15264350 36 16734781946

7 19 2894799 17 25329 27 30859444 37 33583213577

8 39 33535839 18 51866 28 62355854 38 67357328359

9 77 381591711 19 106427 29 125773168 39 135056786787

10 168 ? 20 217216 30 253461052 40 ?

Table 1: Numbers of dismal primes with kdigits in bases 2 and 10 (A169912, A087636).

We can, of course, study the primes dividing n, even if ndoes not contain a digit equal

to β. Without giving any details, we mention that [17] contains the following sequences: the

number of distinct prime divisors of n(A088469), their dismal sum (A088470), and dismal

product (A088471);2also the lists of numbers nsuch that the dismal sum of the distinct

prime divisors of nis < n (A088472), ≤n(A088473), ≥n(A088475), > n (A088476); as

well as the numbers nsuch that the dismal product of the distinct prime divisors of nis < n

(A088477), ≤n(A088478), = n(A088479), ≥n(A088480), and > n (A088481). There is

no analogue of A088476 or A088481 in ordinary arithmetic.

One omission from the above list is explained by the following theorem.

2A088471 has an unusual beginning: 9,9,9,9,9,9,9,9,9,90,123456789987654321,19,19,19, . . . .

8

Theorem 9. In base bdismal arithmetic, nis prime if and only if the dismal sum of its

distinct dismal prime divisors is equal to n.

Proof. If n=pis prime then the sum of the primes dividing it is p. Suppose nis not prime

and let mbe the sum of the distinct dismal primes diving n. If nis divisible by a prime

pwith lenb(p) = lenb(n), then n=rbp,r < β, the largest digit in nis r, and so m6=n

since n≪bp≪bm. If lenb(p)<lenb(n) for all prime divisors p, then lenb(m)<lenb(n), and

again m6=n.

We now consider how many primes there are. Let πb(k) denote the number of base b

dismal primes with kdigits. Table 1shows the initial values of π2(k) and π10(k). Necessary

conditions for a number nto be prime are that it contain βas a digit and (if k > 2) does

not end with 0. There are

(b−1)2bk−2−(b−2) (b−1)k−2(15)

such numbers. It seems likely that, as kincreases, almost all of these numbers will be prime,

and the data in Table 1is consistent with this. We therefore make the following conjecture.

Conjecture 1.

πk(b)∼(b−1)2bk−2as k→ ∞ .(16)

We can get a lower bound on πb(k) by producing large numbers of primes, using the

process of “promotion.” We call a base bnumber with at least two digits a pseudoprime if

its only factorizations are of the form n=pbqwhere at least one of pand qhas length

1. In base b,nis a prime if and only if it is a pseudoprime and contains a digit β. If

nk−1nk−2...n0|bis a pseudoprime and ris its maximal digit, then nk−1nk−2...n0|r+1 is a base

r+ 1 prime and furthermore nk−1nk−2...n0|cis a pseudoprime in any base c≥r+ 1. In base

2 there is no diﬀerence between primes and pseudoprimes. As long as we exclude numbers

ending with 0, reversing the digits of a number does not change its status as a prime or

pseudoprime.

The advantage of working with pseudoprimes rather than primes is that the inverse image

of a pseudoprime under a digit map (see §2) is again a pseudoprime.

Theorem 10. For a base bdigit map g, if g(nk−1nk−2...n0|b)is a pseudoprime and g(nk−1)

is not 0, then nk−1nk−2...n0|bis a pseudoprime.

Proof. This follows immediately from Theorem 3.

So if pis a pseudoprime, then any number nwith the property that there is a digit map

sending nto pis also a pseudoprime; we think of nas being obtained by “promoting” p, and

call pthe “template” for n.

Here is an equivalent way to describe the promotion process. Suppose the distinct digits

in p, the template, or number to be promoted, are d1< d2< d3,.... For each di, choose a

set of digits S(di) such that all the digits in S(di) are strictly less than all those in S(dj),

for i < j. Replace any digit diin pby any digit in S(di). All numbers nobtained in this

way are promoted versions of p(the required digit map gbeing deﬁned by g(c) = difor all

c∈S(di)).

9

Any pseudoprime (in any base) with at most four digits can be obtained by promoting

a base 2 prime. At length 2, 11|2is a prime (see (12)), so every 2-digit number rs|bis a

pseudoprime, using the digit map gthat sends rand sto 1. This is valid even if s= 0,

since (8) still holds. At length 3, 101|2is a base 2 prime, so any three-digit number rst|b

with r > s,t > s is a pseudoprime (take g(r) = g(t) = 1, g(s) = 0), and this captures all

three-digit pseudoprimes. There are three templates of length 4, 1001|2, 1011|2, and 1101|2.

These can be promoted to capture all four-digit primes, which are the numbers rstu|bfor

which one of the following holds:

rand uare both strictly greater than sand t ,

r, s, and uare strictly greater than t ,

r, t, and uare strictly greater than s .

For lengths greater than four, we must use some nonbinary templates to capture all

pseudoprimes, and as the length kincreases so does the fraction of nonbinary templates

required, as shown in Table 2. The columns labeled (a) and (b) give the number of binary

templates and the total number of templates, respectively, and the columns (c) and (d) give

the number of base 10 primes obtained by promoting the templates in columns (a) and (b).

k(a) (b) (c) (d)

2 1 1 18 18

3 1 1 81 81

4 3 3 1539 1539

5 5 8 17661 20457

6 9 51 135489 242217

Table 2: For lengths k= 2 through 6, the numbers of (a) binary templates, (b) all templates,

(c) base 10 primes obtained by promoting the binary templates, and (d) base 10 primes

obtained by promoting all the templates.

We can reduce the list of templates by omitting those that are reversals of others. Table

3shows the reduced list of templates of lengths ≤6.

Since 1 00 ...0

|{z }

k−2

1|2is prime, the promotion process tells us for example that the numbers

r s1s2. . . sk−2

|{z }

k−2

β|bare prime, for 1 ≤r≤β, provided each of the digits siis in the range 0

through r−1. This gives (b−1)k−2+ 2(b−2)k−2+··· =O(bk−2) primes of length k, which

for b > 2 is of exponential growth but smaller than (16).

4 Dismal squares

One might expect that it would be easier to ﬁnd the number of dismal squares of a given

length than the number of dismal primes, but we have not investigated squares as thoroughly,

10

11 100001 102212 120212

101 100011 102221 120221

1001 100101 103223 120222

1011 100111 103233 121022

10001 101011 110212 121102

10011 101221 112021 122102

10111 101222 112022 122202

12021 102201 120021 132023

12022 102202 120022 133023

Table 3: Reduced list of prime templates: every pseudoprime of length ≤6 can be obtained

by promoting one of these 36 pseudoprimes or its reversal (A191420).

and we do not even have a precise conjecture about their asymptotic behavior. In base 2,

the ﬁrst few dismal squares, written in base 10, are

0,1,4,7,16,21,28,31,64,73,84,95,112,125,124,127,256,273, . . . , (17)

(A067398, also contributed by Jens Voß in 2002), and the numbers of squares of lengths 1

(including 0), 3,5,7,... are

2,2,4,8,15,29,55,105,197,367,678,1261,2326,4293,7902,14431, . . . (18)

(A190820). In base 10, the ﬁrst few squares were given in (2), and the numbers of squares

of lengths 1 (including 0), 3, 5, 7,... are

10,90,900,9000,74667,608673,...

(A172199). The sequence of base 10 squares is not monotonic (for example 1011 <1020

yet 1011 10 1011 = 1011111 >1020 10 1020 = 1010200), and contains repetitions. The

numbers which are squares in more than one way are

111111111,111111112,111111113,111111114,111111115,111111116,111111117, . . . ,

e.g., 111111111 = 11011 10 11011 = 11111 10 11111 (A180513, A181319).

We brieﬂy mention two other questions about squares to which we do not know the

answer: (i) In base 2, how many square roots does 22k+1 −1 have? This is a kind of

combinatorial covering problem. For k= 0,1,... the counts are

1,1,1,1,2,3,5,9,15,28,50,95,174,337,637,1231,2373,4618,8974,17567,34387,

67561,132945,262096,517373,1023366,2025627,4014861,7964971,15814414,

31424805,62490481,124330234,247514283,492990898,982307460,1958093809,

3904594162,7788271542,15539347702,31012331211, . . . (19)

(A191701). Is there a formula or recurrence for this sequence? (ii) In base b, if we consider

all psuch that pbp=n, does one of them dominate all the others (in the ≫bsense)? If so

the dominating one could be called the “principal” square root.

11

5 The divisibility poset

One drawback to dismal arithmetic is that there is more than one way to order the dismal

numbers, and no ordering is fully satisfactory.

The usual order on the nonnegative integers (<or ≤) is unsatisfactory, since (working

in base 10) we have 18 <25 yet 18 32 = 38 >25 32 = 35, 32 <41 yet 32 3 = 32 >

41 3 = 31.

The dominance order (≪b) is more satisfactory, in view of Lemma 6and the distributive

law of Theorem 1, but has the drawback that mdivides ndoes not imply that m≪bn(e.g.,

12|10 divides 11|10, yet 11|10 ≪10 12|10).

The partial order induced by divisibility (≺b) is worth discussing, as it has some interest-

ing properties and is the best way to look at dismal numbers as long as we are considering

only questions of factorization and divisibility. For simplicity we will restrict the discussion

to base 10.

...

3

4

7

8

9

10

11

3

1

8

7

6

3

2

16

88

10

20

30

60

70

11

12 22 21

3132332313

616465665646

717275767767572717

8081828386877868382818

... ... ... ... ... ... ... ... ... ... ...

...

...

949919 9091929397988979493929

9

8

5

1

2

7

6

...

... ...

...

...

...

...

...

...

...

...

..................

0

1

2

1100

111

1000

110

11101111

100

Figure 1: Beginning of the divisibility poset (see text for description). The left-hand column

gives the rank (A161813).

Figure 1displays the beginning of the Hasse diagram ([21, p. 99]) of this partially ordered

set (or poset), and shows all positive numbers with one or two digits, and a few larger

numbers. There are too many edges to draw in the diagram, so we will describe them in

words.

12

The multiplicative identity 9, the “zero element” in the poset, is at the base. The other

single-digit numbers 8 ≺7≺6≺ · · · ≺ 1 are above it in the left-hand column, and 8 is

joined to 9. The numbers are arranged in rows according to their rank (shown at the extreme

left of the diagram). The numbers of rank 1 consist of 8 and the (inﬁnitely many) primes:

19,29,...,91,90,109,209,219,309, . . . (A144171). All of these are joined to 9. The numbers

of rank 2 are 7,18,28,...,81,80,108,119,...(A144175), and so on. Every two-digit number

of rank h(1 ≤h≤9) is joined to the single-digit number h−1 on the left of the diagram.,

as indicated by the square brackets. A two-digit number to the left of the central column of

the pyramid is joined to the number diagonally below it to the left (e.g., 56 is joined to 57),

and a two-digit number to the right of the central column is joined to the number diagonally

below it to the right (e.g., 65 is joined to 75). A number in the central column is joined

to the three numbers immediately below it (e.g., 66 is joined to 67,77,76). A number r0

(1 ≤r≤9) in the right-hand column is joined to the number immediately below it and to

the single-digit number r.

Only a few numbers with more than two digits are shown, but a more complete diagram

would show for example that 1 is joined to, besides 10 and 11, many other decimal numbers

whose digits are 0’s and 1’s, such as 101, 1001, 1011, . . .. all of rank 9. The ﬁgure is complete

in the sense that all downward joins are shown for all the numbers in the diagram.

One perhaps surprising property of the divisibility poset is that the greatest lower bound

(or greatest common divisor) m∧nand the least upper bound (or least common multiple)

m∨nof two dismal numbers mand nneed not exist, and so this poset fails to be a lattice

[9, Chap. 1]. For example, again working in base 10, the rank 2 numbers 8989 and 9898

are each divisible by (and joined to) the nine primes 909,919,...,989. However, these nine

primes are incomparable in the ≺10 order, so neither 8989 ∧9898 nor 909 ∨919 exist.

Although greatest common divisors need not exist, we can still deﬁne two dismal numbers

to be relatively prime if their only common divisor is the unit β.

In the next section we will study the number of divisors function db(n). Candidates for

divisors of nare all numbers mwith lenb(m)≤lenb(n). It is therefore appropriate to deﬁne

the dismal analogue of the Euler totient function, φb(n), to be the number of numbers m

with lenb(m)≤lenb(n) which are relatively prime to n. The initial values of φ2(n) and φ10(n)

are shown in Table 4.

n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

φ2(n) 1 2 2 4 6 2 4 8 14 6 14 5 14 5 7 16 30 14 30 12

φ10(n) 1 1 1 1 1 1 1 1 9 18 2 18 18 18 18 18 18 18 90 18

Table 4: Values of totient functions φ2(n) and φ10(n) (A191674, A191675).

6 The number of dismal divisors

Let db(n) denote the number of dismal divisors of nin base b, and let σb(n) denote the

dismal sum of the dismal divisors of n. These functions are more irregular than their classical

13

analogues, as can be seen from the examples in Table 5, and there are no simple formulas for

them. In this section we study some of the properties of db(n). Note that if ris the largest

digit in n, then the smallest divisor of nis r, and the largest divisor is the number obtained

by changing all the r’s in nto β’s.

ndivisors (base 10) d10(n)σ10(n)

1 1,2,3,4,5,6,7,8,9 9 9

2 2,3,4,5,6,7,8,9 8 9

3 3,4,5,6,7,8,9 7 9

4 4,5,6,7,8,9 6 9

5 5,6,7,8,9 5 9

6 6,7,8,9 4 9

7 7,8,9 3 9

8 8,9 2 9

9 9 1 9

10 1,...,9,10,20,...,90 18 99

11 1,...,9, rs with 1 ≤r, s ≤9 90 99

12 2,...,9,12,13,...,19 16 19

Table 5: In base 10, the dismal divisors of the numbers 1 through 12 and the corresponding

values of d10(n) and σ10(n) (A189506, A087029, A087416).

A base bdismal prime phas two divisors, b−1 and p, so db(p) = 2. In the other direction,

a divisor of a k-digit number nhas at most kdigits, so

2≤db(n)≤bk−1.(20)

Base bnumbers of the form 111 ...1|b(that is, in which all the base bdigits are 1) come

close to meeting this upper bound—see Remark (iv) following Theorem 13. We make the

following conjectures.

Conjecture 2. In any base b≥3, among all k-digit numbers n,db(n)has a unique maximum

at n= (bk−1)/(b−1) = 111 ...1|b.

Conjecture 3. In base 2, among all k-digit numbers n, the maximal value of d2(n)occurs

at n= 2k−2 = 111 ...10|2, and this is the unique maximum for n6= 2,4.

Conjecture 4. In base 2, among all odd k-digit numbers n,d2(n)has a unique maximum

at n= 2k−1 = 111 ...111|2, and if k≥3and k6= 5, the second-largest value of d2(n)occurs

at n= 2k−3 = 111 ...101|2,n= 2k−2k−2−1 = 101 ...111|2, and possibly other values of

n.

The numerical evidence supporting these conjectures is compelling. For example, in base

10, if we study the sequence d10(n), n≥1 (A087029) for n≤106, and write down d10(n)

each time it exceeds d10(m) for all m < n, we obtain the values

9,18,90,180,819,1638,7461,14922,67968 ,

14

(see A186443) at these (decimal) values of n:

1,10,11,110,111,1110,1111,11110,11111 .

If nis a 5-digit decimal number, the eight largest values of d10(n) are, in decreasing order,

67968,39624,21812,14922,11202,9616,6732,6570,

at these values of n:

11111,22222,33333,11110,44444,12222 or 22220 or 22221,11011,10111 or 11101.

The number n= (10k−1)/9 = 111 ...1|10 is a clear winner among all k-digit decimal numbers

for k≤5. The data for bases 3 through 9 is equally supportive of Conjecture 2. Likewise, the

binary data strongly supports Conjectures 3and 4—see Table 6for the initial values of d2(n).

Table 6also shows why k= 5 is mentioned as an exception in Conjecture 4: among 5-digit

odd numbers, d2(11011|2) = 4 is the runner-up, ahead of d2(10111|2) = d2(11101|2) = 2.

n(base 2) d2(n)n(base 2) d2(n)n(base 2) d2(n)n(base 2) d2(n)

− − 1000 4 10000 5 11000 8

1 1 1001 2 10001 2 11001 2

10 2 1010 4 10010 4 11010 4

11 2 1011 2 10011 2 11011 4

100 3 1100 6 10100 6 11100 9

101 2 1101 2 10101 3 11101 2

110 4 1110 6 10110 4 11110 10

111 3 1111 5 10111 2 11111 8

Table 6: In base 2, the number of dismal divisors of the numbers 1 through 31 (A067399).

For a more dramatic illustration of Conjectures 2and 3, see the graphs of sequences

A087029 and A067399 in [17]. Although the evidence is convincing, we have not, unfortu-

nately, succeeded in proving these conjectures.

We are able to determine the exact values of db(111 ...111|b) for all b(the conjectural

winner for b≥3 and the conjectural winner among odd numbers in the binary case) and

d2(111 ...101|2) = d2(101 ...111) (the conjectural runners-up among odd binary numbers of

length k > 5).

We begin with a lemma that describes the eﬀect of trailing zeros.

Lemma 11. If the base bexpansion of nends with exactly r≥0zeros, so that n=mbr,

with b∤m, then

db(n) = (r+ 1)db(m).

Proof. If pbq=mthen b∤p,b∤q, and the r+ 1 numbers pbi(0 ≤i≤r) dismally divide

n, since

(pbi)b(qbr−i) = n .

Conversely, if p′bq′=nthen p′=pbi,q′=qbr−i,b∤p,b∤q, for some iwith 0 ≤i≤r. So

each dismal divisor of mcorresponds to exactly r+ 1 dismal divisors of n.

15

For example, in base 10 the dismal divisors of 7 are 7,8,9 and the dismal divisors of 700

are 7,8,9,70,80,90,700,800,900.

One reason Conjectures 2-4seem hard to prove is the erratic behavior of db(n). In contrast

to the above lemma, the eﬀect of internal zeros is hard to analyze. Suppose the i-th digit in

the b-ary expansion of nis zero. This implies that if n=pbq, then all entries in the i-th

column of the long multiplication tableau must be zero, which imposes many constraints on

the b-ary expansions of pand q. One would expect, therefore, that changing the zero digit

to a larger number—thus weakening the constraints—would always increase the number of

divisors of n. Roughly speaking, this is true, but there are many cases where it fails. For

example, d2(11111|2) = 8, but d2(11110|2) = 10 (see Table 6, Lemma 11 and Theorem 12).

Again, d2(10101|2) = 3, but 10111|2is prime, so d2(10111|2) = 2. In any base b, 11 00 ...0

|{z }

r

|b

has (r+ 1)((b−1)2+ (b−1)) divisors, whereas 11 00 ...0

|{z }

r−1

1|bhas (b−1)3+ (b−1) divisors,

a smaller number if ris large.

The next result was conjectured by the second author and proved by Richard Schroeppel

in 2001 [15]. An alternative proof (via a bijection with a certain class of polyominoes) was

given by Frosini and Rinaldi in 2006 [6]. We give a version of Schroeppel’s elegant direct

proof, partly because it has never been published, and partly because we will use similar

arguments later.

Theorem 12. In base 2, the number of dismal divisors of 111 ...1|2(with k1’s)is equal to

the number of compositions of kinto parts of which the ﬁrst is at least as great as all the

other parts.

Proof. Suppose p2q= 111 ...1|2(with k1’s) where len2(p) = r, len2(q) = k+ 1 −r.

By examining the long multiplication tableau for p2q, we see that it is also true that

p2q′= 111 ...1|2where q′= 111 ...1|2, with k+ 1 −r1’s (for if there is a 1 in each column

of the tableau for p2q, that is still true for p2q′). So in order to ﬁnd all the divisors p

of 111 ...1|2(with k1’s) we may assume that the cofactor qhas the form 111 ...1|2(with s

1’s, for some s, 1 ≤s≤k).

We establish the desired result by exhibiting a bijection between the two sets. Let

k=c1+c2+...+ctbe a composition of kin which c1≥ci≥1 (2 ≤i≤t). Let ψ(ci) denote

the binary vector 000 ...01 with ci−1 0’s and a single 1. The divisor pcorresponding to

this composition has binary representation given by the concatenation

1ψ(c2)ψ(c3). . . ψ(cr)|2,(21)

of length k+ 1 −c1. If we set q= 111 ...1|2, of length c1, then p2q= 111 ...1|2(with

k1’s). This follows from the fact that if the binary representation of pcontains a string of

exactly s0’s:

...1 000 ...0

|{z }

s

1. . . ,

then, when we form the product p2q, the 1 immediately to the right of these 0’s will

propagate leftward to cover the 0’s if and only if len2(q)≥s+ 1, which is exactly the

condition that c1≥cifor all i≥2.

16

For example, the eight compositions of 5 in which no part exceeds the ﬁrst and the

corresponding factorizations p2qof 11111|2are shown in Table 7. Dots have been inserted

in pto indicate the division into the pieces ψ(ci).

Composition of 5 divisor pcofactor q

5 1|211111|2

41 1.1|21111|2

32 1.01|2111|2

311 1.1.1|2111|2

221 1.01.1|211|2

212 1.1.01|211|2

2111 1.1.1.1|211|2

11111 1.1.1.1.1|21|2

Table 7: Illustrating the bijection used to prove Theorem 12.

Remarks.

(i) Using the bijection deﬁned by (21), the number of 1’s in the binary expansion of the

divisor pis equal to the number of parts in the corresponding composition.

(ii) It follows immediately from the interpretation in terms of compositions that the

numbers d2(2k−1) have generating function

∞

X

k=1

d2(2k−1) zk=

∞

X

l=1

zl

1−(z+z2+···+zl)

=

∞

X

l=1

(1 −z)zl

1−2z+zl+1 (22)

(the index of summation, l, corresponds to the ﬁrst part in the composition).

(iii) The initial values of this sequence are shown in Table 8. This sequence appears in

k1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

d2(2k−1) 1 2 3 5 8 14 24 43 77 140 256 472 874 1628 3045 5719

Table 8: Values of d2(2k−1).

entries A007059 and A079500 in [17], although the indexing is diﬀerent in each case. The

sequence also occurs in at least four other contexts besides the two mentioned in Theorem

12, namely in the enumeration of balanced ordered trees (Kemp [10]), of polyominoes that

tile the plane by translation (Beauquier and Nivat [2], Brlek et al. [3]), of Dyck paths (see

A007059), and in counting solutions to the postage stamp problem (again see A007059).

The article by Frosini and Rinaldi [6] gives bijections between four of these six enumerations.

In this context we should also mention the recent article of Rawlings and Tiefenbruck [19],

17

which, although not directly related to the problems we consider, discusses other connections

between the enumeration of compositions, permutations, polyominoes, and binary words.

(iv) The asymptotic behavior of this sequence is quite subtle. From the work of Kemp

[10] and Knopfmacher and Robbins [12] it follows that

d2(2k−1) ∼2k

klog 2 (1 + Θk),as k→ ∞ ,(23)

where Θkis a bounded oscillating function with |Θk|<10−5(see the proof of Theorem 19

below).

In order to determine db(111 ...1|b) for bases b > 2, we ﬁrst classify compositions in

which no part exceeds the ﬁrst according to the number of parts. Let T(k, t) denote the

number of compositions of kinto exactly tparts (with 1 ≤t≤k) such that no part exceeds

the ﬁrst. Table 9shows the initial values. This is entry A184957 in [17].3

k\t1 2 3 4 5 6 7 8

1 1

2 1 1

3 1 1 1

4 1 2 1 1

5 1 2 3 1 1

6 1 3 4 4 1 1

7 1 3 6 7 5 1 1

8 1 4 8 11 11 6 1 1

Table 9: Initial values of of T(k, t), the number of compositions of kinto exactly tparts such

that no part exceeds the ﬁrst.

The values of T(k, t) are easily computed via the auxiliary variables γ(k, t, m), which we

deﬁne to be the number of compositions of kinto tparts of which the ﬁrst part, m, is the

greatest (for 1 ≤t≤k, 1 ≤m≤k). We have the recurrence

γ(k, t, m) =

min{m,k+2−t−m}

X

j=1

γ(k−j, t −1, m) (24)

(classifying compositions according to the last part, j), for m > 1, t > 1, t+m < k −1, with

initial conditions

γ(k, t, 1) = δt,k ,

γ(k, 1, m) = δm,k ,

3An array equivalent to this, A156041, was contributed to [17] by J. Grahl in 2009 and later studied by

A. P. Heinz and R. H. Hardin.

18

where δi,j = 1 if i=jor 0 if i6=j. Then

T(k, t) =

k+1−t

X

m=1

γ(k, t, m).(25)

Since γ(k, t, m) is the coeﬃcient of zkin zm(z+z2+···+zm)t−1, it follows that column t

of Table 9has generating function

∞

X

k=1

T(k, t)zk=zt−1

(1 −z)t−1

t−1

X

r=0

(−1)rt−1

rzr+1

1−zr+1 .(26)

Since the total number of compositions of kinto tparts is k−1

t−1, and in at least a fraction

1

tof them the ﬁrst part is the greatest, we have the bounds

1

tk−1

t−1≤T(k, t)≤k−1

t−1.(27)

Theorem 13.

dbbk−1

b−1=db(11 ...1

|{z }

k

|b) =

k

X

t=1

T(k, t)(b−1)t.(28)

Proof. Suppose pbq= 11 ...1

|{z }

k

|b. At least one of pand q, say q, must contain only digits 0

and 1 (for if pcontains a digit i > 1 and qcontains a digit j > 1, then ibj= min{i, j }>1

will appear somewhere in pbq). As in the proof of Theorem 12 we may assume that this

qhas the form 11 ...1

|{z }

s

|bfor some swith 1 ≤s≤k. Suppose p=Pr−1

i=0 pi2iwith pi∈ {0,1}

is a divisor of 11 ...1

|{z }

k

|2, so that

p2(2k+1−r−1) = 2k−1.(29)

By Lemma 5,p′:= Pr−1

i=0 pibiis a base bdismal divisor of bk−1

b−1:

p′b

bk+1−r−1

b−1=bk−1

b−1.(30)

Furthermore, (30) still holds if any of the pithat are 1 are changed to any digit in the range

{1,2,...,b−1}. Conversely, any base bdismal divisor p′of bk−1

b−1remains a divisor if all the

nonzero digits in the base bexpansion of p′are replaced by 1’s. So each divisor of 11 ...1

|{z }

k

|2

with t1’s corresponds to (b−1)tdivisors of bk−1

b−1. Since there are T(k, t) divisors of 2k−1

with t1’s, the result follows.

Remarks.

(i) Table 10 shows the initial values of db(11 ...1

|{z }

k

|b).

19

k\b2 3 4 5 6 7 8 9 10

1 1 2 3 4 5 6 7 8 9

2 2 6 12 20 30 42 56 72 90

3 3 14 39 84 155 258 399 584 819

4 5 34 129 356 805 1590 2849 4744 7461

5 8 82 426 1508 4180 9798 20342 38536 67968

6 14 206 1434 6452 21830 60594 145586 313544 619902

7 24 526 4890 27828 114580 375954 1044246 2555080 5660208

Table 10: Table of db(11 ...1|b) (with k1’s), the number of base bdismal divisors of bk−1

b−1

(rows: A002378, A027444, A186636; columns A079500, A186523).

(ii) Theorem 13 reduces to Theorem 12 in the case b= 2.

(iii) From (27) and (28) we have

bk−1

k≤db(11 ...1

|{z }

k

|b)≤(b−1)bk−1.(31)

For b= 2 we also have the asymptotic estimate (23).

We now study the runners-up in the binary case (among odd numbers of length greater

than 5), namely the numbers 111 ...101|2and 101 ...111|2. The simplest way to state the

result is to give the generating function.

Theorem 14.

∞

X

k=3

d2(2k−3)zk=z+z3

1−z+

∞

X

l=3

(1 −z)2zl

1−2z+zl−1−zl+zl+2

=z+ 2z3+ 2z4+ 2z5+ 4z6+ 6z7+ 10z8+··· .(A188288) (32)

We will deduce Theorem 14 from Theorem 17 below.

Suppose p2q= 2k−3, k≥3, where len2(p) = h, len2(q) = l, with h+l=k+ 1. In

order to ﬁnd all choices for p, we note that the binary expansions of pand qmust end with

...01, and that, as in the proofs of Theorems 12 and 13, we may assume that q= 2l−3.

Our approach is to ﬁx land allow hto vary. Let M(l)

hdenote the number of binary numbers

pwith len2(p) = hsuch that

p2111 ...101

|{z }

l

|2= 111 ...101

|{z }

h+l−1

|2.(33)

Suppose the binary expansion of pis

1xvxv−1. . . x3x2x10 1|2,

where v:= h−3 and the xiare 0 or 1. The long multiplication tableau for (33) implies that

the ximust satisfy certain Boolean equations (remember that 2is the logical OR; in what

follows we will write rather than 2). For example, the tableau for l= 4 and h= 9,

v= 6 is shown in Figure 2.

20

1x6x5x4x3x2x10 1

21 1 0 1

1x6x5x4x3x2x10 1

1x6x5x4x3x2x10 1

1x6x5x4x3x2x10 1

111111111101

Fig. 2.

By reading down the columns, we obtain the equations

x1x3= 1 ,

x1x2x4= 1 ,

x2x3x5= 1 ,

x3x4x6= 1 ,

x5x6= 1 .

There are M(4)

9= 29 solutions (x1,...,x6) to these equations. Table 11 shows the initial

values of M(l)

h, as found by computer.

h\l1 2 3 4 5 6 7 8

1 1 0 1 1 1 1 1 1

2 0 0 0 0 0 0 0 0

3 1 0 0 1 1 1 1 1

4 1 0 1 1 2 2 2 2

5 1 0 2 3 3 4 4 4

6 1 0 2 5 7 7 8 8

7 1 0 3 9 13 15 15 16

8 1 0 6 16 24 29 31 31

9 1 0 10 29 47 56 61 63

10 1 0 15 53 89 110 120 125

11 1 0 24 96 170 216 238 248

12 1 0 40 174 326 422 471 494

Table 11: Table of M(l)

h(columns 3 and 4 are A070550 and A188223).

Inspection of the table suggests that the l-th column satisﬁes the recurrence

M(l)

h=M(l)

h−1+M(l)

h−2+···+M(l)

h−l+2 +M(l)

h−l+M(l)

h−l−1,(34)

for l≥3. This will be established in Corollary 18.

We consider the cases h≤l+1 and h≥l+2 separately. For h≤l+1, it is straightforward

to show the following:

M(1)

2= 0 , M (1)

h= 1 (h6= 2) , M (2)

h= 0 ,(35)

21

and, for l≥3, h≤l+ 1,

M(l)

h=

1,if h= 1,

0,if h= 2,

2h−3,if 3 ≤h≤l−1,

2h−3−1,if h=lor l+ 1.

(36)

This accounts for the entries in Table 11 that are on or above the line h−l= 1.

We now consider the case 3 ≤l≤h−2 = v+ 1. The multiplication tableau leads to two

special equations, (x2= 1,if l= 3, or

x1x2··· xl−3xl−1= 1,if l≥4,(37)

and

xv−l+3 xv−l+4 ··· xv−1xv= 1 ,(38)

together with a family of v−l+ 1 further equations, which, if l= 3, are

x1x3=x2x4=··· =xv−3xv−1=xv−2xv= 1 ,(39)

or, if l≥4, are

x1x2x3··· xl−2xl= 1 ,

x2x3x4··· xl−1xl+1 = 1 ,

...............

xv−l+1 xv−l+2 xv−l+3 ··· xv−2xv= 1 .

(40)

The two special equations (37) and (38) involve variables with both low and high indices,

which makes induction diﬃcult. We therefore deﬁne a simpler system of Boolean equations

in which the special constraints apply only to the high-indexed variables.

For l≥3 and n≥1, let D(l)

ndenote the number of binary vectors x1x2...xnof length n

that end with xn= 1, do not contain any substring

00 ...000

|{z }

l

or 00 ...010

|{z }

l

and do not end with

00 ...01

|{z }

l−1

.

Equivalently, D(l)

nis the number of solutions to the Boolean equations

x1x2x3··· xl−2xl= 1 ,

x2x3x4··· xl−1xl+1 = 1 ,

...............

xn−lxn−l+1 xn−l+2 ··· xn−3xn−1= 1 ,(41)

22

and

xn−l+2 xn−l+3 ··· xn−2xn−1= 1 , xn= 1 .(42)

We also set D(l)

0= 1.

Theorem 15. For l≥3,n≥1, there is a one-to-one correspondence between binary vectors

of length nsatisfying the D(l)

nequations and compositions of ninto parts taken from the set

{1,2,...,l−2, l, l + 1}.(43)

Proof. We exhibit a bijection between the two sets. Let cbe a composition n=c1+c2+···+cr

into parts from (43). For 1 ≤i≤l−2, let ψ(i) = 00 ...01, of length iand ending with a

single 1, let ψ(l) = 00 ...011, of length l, let ψ(l+ 1) = 00 ...0011, of length l+ 1, and let

ψ(c) = ψ(c1)ψ(c2)··· ψ(cr),(44)

a binary vector of length n. Note that the ψ(i) for 1 ≤i≤l−2 contain runs of at most l−3

zeros. Runs of l−2 or l−1 zeros in ψ(c) are therefore followed by two ones. So conditions

(41) and (42) are satisﬁed. Conversely, given a binary vector satisfying the D(l)

nequations,

we can decompose it into substrings ψ(i) by reading it from left to right.

n\l3 4 5 6

0 1 1 1 1

1 1 1 1 1

2 1 2 2 2

3 2 3 4 4

4 4 6 7 8

5 6 11 14 15

6 9 20 27 30

7 15 36 51 59

8 25 65 98 115

Table 12: Table of D(l)

n(the columns are A006498, A079976, A079968, A189101).

The generating function for the D(l)

nfollows immediately from the theorem:

Corollary 16. For l≥3, the numbers D(l)

nhave generating function

D(l)(z) :=

∞

X

n=0

D(l)

nzn=1

1−(z+z2+···+zl−2+zl+zl+1 )

=1−z

1−2z+zl−1−zl+zl+2 .(45)

23

Table 12 shows the initial values of D(l)

n, computed using the generating function. The

l= 4 and l= 5 columns are in [17] as entries A079976 and A079968, taken from a paper by D.

H. Lehmer on enumerating permutations (π1,...,πn) with restrictions on the displacements

πi−i([14]; see also [11]).

We now express the numbers M(l)

hin terms of the D(l)

n. We consider the lvalues x1,...,xl

in a solution to the M(l)

hequations, and classify them according to the number of leading

zeros. There are just l−1 possibilities, as shown in Table 13, and in each case the M(l)

hequa-

tions reduce to an instance of the D(l)

nequations. For example, if x1= 1 the M(l)

hequations

reduce to an instance of the D(l)

h−3equations. (E.g., if l= 5 and we set x1= 1, equations (37),

(38), (40) become xv−2xv−1xv= 1, x2x3x4x6= 1, ...,xv−4xv−3xv−2xv= 1,

which, if we subtract 1 from each subscript, are the equations for D(5)

v, that is, D(5)

h−3.)

Setting, in M(l)

h,leads to

x1= 1 D(l)

h−3

x1= 0, x2= 1 D(l)

h−4

x1=x2= 0, x3= 1 D(l)

h−5

· · · · · ·

x1=···=xl−4= 0, xl−3= 1 D(l)

h−l+1

x1=···=xl−3= 0, xl−2=xl−1= 1 D(l)

h−l−1

x1=···=xl−2= 0, xl−1=xl= 1 D(l)

h−l−2

Table 13: Expressing M(l)

hin terms of D(l)

n.

We have therefore shown that for l≥3, h≥3,

M(l)

h=D(l)

h−3+D(l)

h−4+D(l)

h−5+···+D(l)

h−l+1 +D(l)

h−l−1+D(l)

h−l−2.(46)

Table 14 illustrates (46) in the case l= 4.

h M(4)

hD(4)

h−3D(4)

h−5D(4)

h−6

3 1 1 − −

4 1 1 − −

5 3 2 1 −

6 5 3 1 1

7 9 6 2 1

8 16 11 3 2

9 29 20 6 3

Table 14: Illustrating M(4)

h=D(4)

h−3+D(4)

h−5+D(4)

h−6.

From (46) and Corollary 16, and taking into account the values of M(l)

hfor h < 3, we

obtain:

24

Theorem 17. For l≥3,

M(l)(z) :=

∞

X

h=1

M(l)

hzh=z+ (z3+z4+···+zl−1+zl+1 +zl+2)D(l)(z)

=z(1 −z)

1−(z+z2+···+zl−2+zl+zl+1)

=z(1 −z)2

1−2z+zl−1−zl+zl+2 .(47)

It is now straightforward to obtain the recurrence for M(l)

hfrom the generating function

in the second line of the display. We omit the proof.

Corollary 18. For l≥3,M(l)

hsatisﬁes the recurrence (34) with initial conditions (35),

(36).

We can now give the proof of Theorem 14. From the deﬁnition of M(l)

h, we have

d2(2k−3) =

k

X

l=1

M(l)

k−l+1 .

That is, d2(2k−3) is the sum of the coeﬃcient of zkin M(1)(z), the coeﬃcient of zk−1in

M(2)(z), ..., and the coeﬃcient of z1in M(k)(z). In other words, d2(2k−3) is the coeﬃcient

of zkin

M(1)(z) + zM(2) (z) + z2M(3)(z) + ···+zk−1M(k)(z),

and now (32) follows from M(1)(z) = z+z3/(1 −z), M(2)(z) = 0, and Theorem 17. This

completes the proof of Theorem 14.

Remark. The l= 3 column of the M(l)

htable (Table 11) is an interesting sequence in its own

right.4To analyze it directly, ﬁrst consider the system of simultaneous Boolean equations

x1x2=x2x3··· xn−1xn= 1 ,(48)

for n≥2, involving a chain of linked pairs of variables. An easy induction shows that the

number of solutions is the Fibonacci number Fn+2 (cf. A000045).5Second, the equations for

M(3)

h, (38) and (39), break up into two disjoint chains like (48), and we ﬁnd that

M(3)

h=(F(n−2)/2Fn/2,if nis even,

F(n−3)/2F(n+1)/2,if nis odd.(49)

4It is entry A070550 in [17], which contains a comment by Ed Pegg, Jr., that it arises in the analysis of

Penney’s game.

5This result could also be obtained by the Goulden-Jackson cluster method, as implemented by Noonan

and Zeilberger [8], [16].

25

From (49) we can derive the recurrence M(3)

h=M(3)

h−1+M(3)

h−3+M(3)

h−4and the generating

function

M(3)(z) = z(1 −z)

1−z−z3−z4,(50)

in agreement with (34) and (47).

The ﬁnal result in this section will describe the asymptotic behavior of the sequence

d2(2k−3). When investigating Conjectures 3and 4, we observed that among all numbers

nwith kbinary digits, the number 2k−1 was the clear winner, with values close to the

estimate (23). The runners-up, a long way behind, were 2k−3 and 2k−2k−2−1 (and

sometimes other values of n), all with the same number of dismal divisors, for which the

number of dismal divisors appeared to be converging to one-ﬁfth of the number of divisors

of the winner, or in other words it appeared that

d2(2k−3)

d2(2k−1) →1

5,as k→ ∞ .(51)

We will now establish this from the generating function (32).

Theorem 19.

d2(2k−3) ∼2k

5klog 2 (1 + ¯

Θk),as k→ ∞ ,(52)

where ¯

Θkis a bounded oscillating function with |¯

Θk|<10−5.

Proof. Our proof is modeled on Knopfmacher and Robbins’s proof [12] of (23), which uses

the method of Mellin transforms as presented by Flajolet, Gourdon, and Dumas [5]. We will

indicate how the Knopfmacher-Robbins proof can be reworded so as to establish (23) and

(52) simultaneously.

Knopfmacher and Robbins work, not with (22), but with

f(z) :=

∞

X

l=1

(1 −z)zl

1−2z+zl,(53)

which is the generating function for the number of compositions of ninto parts of which the

ﬁrst is strictly greater than all the other parts (A007059). Equations (22) and (53) basically

diﬀer just by a factor of z. Then [12] shows that the coeﬃcient of znin f(z) is

2n−1

nlog 2(1 + Θ) ,(54)

for some small oscillating function Θ, which implies (23).

So as to have a function with the same form as (53), we consider, not (32), but

f(z) :=

∞

X

l=2

(1 −z)2zl

1−2z+zl−zl+1 +zl+3 ,(55)

26

and we will show that the coeﬃcient of znis

2n+1

5nlog 2(1 + Θ) ,(56)

for some (diﬀerent) small oscillating function Θ, which implies (52). We can change the

lower index of summation in (55) from 2 to 1, since the l= 1 term is the generating function

for the Padovan sequence (A000931), which grows at a much slower rate than (56).

In what follows, we simply record how the expressions in Knopfmacher and Robbins’s

proof [12] need to be modiﬁed so as to apply simultaneously to (53), which we refer to as case

I, and (55), which we call case II. We follow Knopfmacher and Robbins’s notation, except

that we use jand mas local variables, rather than k, to avoid confusion with the kin the

statement of the theorem. Some typographical errors in [12] have been silently corrected.

Let ρjdenote the smallest root of the denominator of the j-th summand in f(z) that lies

between 0 and 1. Then

ρj=1

21 + τ2−j+O(j2−2j),(57)

where τ= 1 (case I) or 5/8 (case II).

Let qn,j denote the coeﬃcient of znin the j-th summand in f(z). Then

qn,j ≈2n−j−ǫ1−τ

2jn

≈2n−j−ǫe−τn/2j,(58)

where ǫ= 1 in case I or 2 in case II. Next, fn, the coeﬃcient of znin f(z), is

2n−ǫ ∞

X

j=2

2−je−τn/2j+o(1)!.

Let

g(x) :=

∞

X

j=2

2−je−τx/2j.

The Mellin transform of g(x) is

g∗(s) := 1

τs

22(s−1)

1−2s−1Γ(s),0<ℜ(s)<1.

To compute fnwe use (following Knopfmacher and Robbins) the Mellin inversion formula

g(x) = 1

2πi Z1/2+i∞

1/2−i∞

x−sg∗(s)ds .

Now g∗(s)x−shas a simple pole at s= 1 + χm, for each m∈Z, where χm= 2πim/ log 2,

with residue

−1

xlog 2

Γ(1 + χm)e−2πim log2x

τ1+χm.

27

After combining the contributions from all the poles, we have

fn= 2n−ǫ

∞

X

m=−∞

1

nlog 2

Γ(1 + 2πim/ log 2) e−2π im log2n

τ1+2πim/ log 2 .(59)

The term for m= 0 dominates, and we obtain the desired results (54) and (56).

7 The sum of dismal divisors

We brieﬂy discuss the dismal sum-of-divisors function σb(n) (see A188548, A190632, and

A087416 for bases 2, 3, and 10).

Theorem 20. In any base b≥2, if lenb(n) = k, then

n≤σb(n)≤bk−1,(60)

and σb(n) = nif and only if n≡b−1 (mod b).

Proof. The ﬁrst assertion follows because ndivides itself, and no divisor has length greater

than k. If n6≡ b−1 (mod b) then since b−1 is the multiplicative unit, σb(n)6=n. Suppose

n≡b−1 (mod b) and pis a divisor of n, with say pbq=n. Both pand qmust end with

β. From the long multiplication tableau, p≪bn, so pbn=n, and therefore σb(n) = n.

In ordinary arithmetic, a number nis perfect if its sum of divisors is 2n. In dismal

arithmetic, nbn=n. So the second part of the theorem might, by a stretch, be interpreted

as saying that the numbers congruent to b−1 (mod b) are the base bperfect dismal numbers.

In base 2, then, σ2(n) = nif and only if nis odd. Examination of the data shows that,

if nis even, with len2(n) = k, often σ2(n) takes its maximal value, 2k−1. Table 15 shows

the ﬁrst few exceptions, which are characterized in the next theorem.

n σ2(n)n σ2(n)

10010 11011 1001010 1101111

100010 110011 1001110 1101111

100110 110111 1010010 1111011

110010 111011 1100010 1110011

1000010 1100011 1100110 1110111

1000100 1110111 1110010 1111011

1000110 1100111 ...... ......

Table 15: Even numbers nsuch that σ2(n) is not of the form 11 ...1|2(A190149–A190151).

Both nand σ2(n) are written in base 2.

Theorem 21. Suppose n= 2rmwith r≥1,modd, and len2(n) = k. Then

σ2(n) = 2k−1 = 11 ...1

|{z }

k

|2

unless the binary expansion of mcontains a run of more than rconsecutive zeros.

28

Proof. Since mis odd, σ2(m) = m. Therefore

σ2(n) = m|22m0|2 2 m00|2 2 ··· 2m00 ...0

|{z }

r

|2,

and any string 00 ...0

|{z }

i

1 in mwill become 11 ...1

|{z }

i

1 in σ2(n) unless iexceeds r.

The ﬁrst entry in Table 15 is explained by the fact that n= 10010|2,r= 1, m= 1001|2,

and mcontains a run of two zeros.

We also considered two other possible deﬁnitions of perfect numbers: (i) nis perfect in

base b≥3 if the dismal sum of the dismal divisors of nis equal to 2 bn. We leave it to

the reader to verify that for this to happen, bmust be 3, and then nis perfect if and only

if n≡2 mod 4. (ii) nis perfect in base b≥2 if the dismal sum of the dismal divisors of n

diﬀerent from nis equal to n. But here ncannot be b−1, so b−1 is a divisor, and nends

with b−1. This implies that nhas no divisors of length lenb(n) except nitself, so the sum

cannot equal n, and therefore no such nexists.

We end this section with a conjecture (see A186442):

Conjecture 5. For all n > 1,d10(n)< σ10(n).

8 Dismal partitions

Since nbn=n, it only makes sense to consider partitions into distinct parts (otherwise

every number has inﬁnitely many diﬀerent partitions). We deﬁne pb(n) to be the number of

ways of writing

n=m1bm2b··· bml,(61)

for some l≥1 and distinct positive integers mi, without regard to the order of summation.

We set pb(0) = 1 by convention.

For example, p3(7) = p3(21|3) = 22, since (working in base 3) 21 is equal to 21 3any

subset of {20,11,10,1}(16 solutions), 20 311 3any subset of {10,1}(4 solutions), and

20 313any subset of {10}(2 solutions), for a total of 22 solutions.

Remarks.

(i) Permuting the digits of ndoes not change pb(n).

(ii) Any zero digits in ncan be ignored. If n′is the base bnumber obtained by dropping

any ni’s that are zero, pb(n′) = pb(n).

(iii) Although we will not make any use of it, there is a generating function for the pb(n)

analogous to that for the classical case. If we interpret zmbznto mean zmbn, then we

have the formal power series

1 + pb(1)z+pb(2)z2+pb(3)z3+··· = (1 + z)b(1 + z2)b(1 + z3)b··· .

(iv) The sequences p2(n) and p10(n) form entries A054244 and A087079 in [17], con-

tributed by the second author in 2000 and 2003, respectively.

In the remainder of this section we index the digits of nby {1,2,...,n}, in order to

simplify the discussion of subsets of these indices.

29

Theorem 22. If n=n1n2. . . nk|2,ni∈ {0,1}, and the binary weight of nis w, then p2(n)

is equal to the number of set-covers of a labeled w-set by nonempty sets (cf. A003465, [4,

p. 165]), that is,

p2(n) = 1

2

w

X

i=0

(−1)w−iw

i22i.(62)

Proof. From Remark (ii), p2(n) = p2(11 ...1

|{z }

w

|2). There is an obvious one-to-one corre-

spondence between collections of distinct nonempty subsets of {1,...,w}whose union is

{1,...,w}and sets of distinct nonzero binary vectors whose dismal sum is 11 ...1

|{z }

w

|2.

Theorem 23. If n=n1n2...nk|b,0≤ni≤b−1, then

pb(n) = 1

2X

S⊆{1,...,k}

(−1)|S|2Qi(ni+ǫi),(63)

where ǫi= 0 if i∈S,ǫi= 1 if i /∈S.

Proof. The set Ωnof x≪bnis a partially ordered set (with respect to the operator ≪b)

with M¨obius function given by [21,§3.8.4]

µ(x1. . . xk|b, y1. . . yk|b) = ((−1)Pi(yi−xi),if yi−xi= 0 or 1 for all i,

0,otherwise.(64)

Every subset of Ωn\ {0}has dismal sum equal to some number ≪bn, so we have

X

x≪bn

pb(x) = 2Qi(ni+1)−1.

From the M¨obius inversion formula [21,§3.7.1] we get

pb(n) = 1

2X

S⊆{1,...,k}

(−1)|S|2Qi∈SniQj /∈S(nj+1) ,

which implies (63).

Theorem 23 reduces to Theorem 22 if all niare 0 or 1.

Corollary 24. Suppose n=n1n2. . . nk|b, and let ybe the ordinary product n1×n2×· · ·×nk.

Then pb(n)is divisible (in ordinary arithmetic!) by 2y−1.

Corollary 25. For a single-digit number n=n1|b,pb(n) = 2n1−1. For a two-digit number

n=n1n2|b,

pb(n) = 2(n1+1)(n2+1)−2−2n1n2−1(2n1−1)(2n2−1) .(65)

Eq. (65) was found by Wasserman [17, entry A087079].

It follows from the above discussion that in any base b, the only numbers nsuch that

pb(n) = 1 are 0|b, 1|b, 10|b, 100|b, 1000|b,..., and that all other numbers nhave the property

that the dismal sum of the numbers x≪bnis n. These two classes might be called “additive

primes” and “additive perfect numbers.”

30

9 Conclusion and future explorations

We have attempted to show that dismal arithmetic, despite its simple deﬁnition, is worth

studying for the interesting problems that arise. We have left many questions unanswered:

the “prime number theorem” of Conjecture 1, the questions about the numbers of divisors

stated in Conjectures 2-4(in particular, is it true that 11 ...1|10 has more divisors than

any other base 10 number with the same number of digits?), the base 10 dismal analog of

d(n)≤σ(n) (Conjecture 5), and the two questions about dismal squares at the end of §4—in

particular, is there a recurrence for the sequence (19)? There are numerous other questions

that we have not investigated (for example, if x≪by, what can be said about xby?).

We have made no mention of the complexity of deciding if a number is a dismal prime,

or of ﬁnding dismal factorizations. In base 2 such questions reduce to solving a set of

simultaneous quadratic Boolean equations, where a typical equation might be

(x02y4)2(x1 2 y3)2··· 2(x4 2 y0) = 0 (or 1) .

This becomes a question about the satisﬁability of a complicated Boolean expression, and is

likely to be hard to solve in general [7].

While we have focused on the cases b= 2 and b= 10, it would be nice to better

understand the qualitative diﬀerences across a wider range of bases. For example, while

b= 2 is a kind of Boolean arithmetic, does b= 3 correspond to a three-valued logic? Do

odd bbehave diﬀerently from even b? More generally, what other interesting mathematical

structures might be modeled by dismal arithmetic?

Acknowledgments

We thank Adam Jobson for computing an extended version of Table 11, which was helpful

in guessing the recurrence (34). Most of the our computations were programmed in C++,

Fortran, Lisp, and Maple. We also made use of Mathematica’s SatisﬁabilityCount command,

and we thank Michael Somos for telling us about it. We are also grateful to Doron Zeilberger

for telling us about his work with John Noonan [16] on implementing the Goulden-Jackson

cluster method. The OEIS [17] repaid us several times during the course of this work by

providing valuable hints, notably from the entries A007059, A067399, A070550, A079500,

A156041, and A164387.

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2010 Mathematics Subject Classiﬁcation: Primary 06A06, 11A25, 11A63; Secondary 11N37.

Keywords: Carryless arithmetic, squares, primes, divisors, compositions, partitions, asymp-

totic expansions, Mellin transform.

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