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Heliyon 8 (2022) e11651

Contents lists available at ScienceDirect

Heliyon

journal homepage: www.cell.com/heliyon

Research article

Generating binomial coeﬃcients in a row of Pascal’s triangle from

extensions of powers of eleven

Md. Shariful Islam a, Md. Robiul Islam b,∗, Md. Shorif Hossan c, Md. Hasan Kibria a

aDepartment of Mathematics, University of Dhaka, Bangladesh

bDepartment of Computer Science and Engineering, Green University of Bangladesh, Dhaka, Bangladesh

cDepartment of Applied Mathematics, University of Dhaka, Bangladesh

A R T I C L E I N F O A B S T R A C T

Keywords:

Binomial coeﬃcients

Pascal’s triangle

Logarithm

Modular arithmetic

Sir Isaac Newton noticed that the values of the ﬁrst ﬁve rows of Pascal’s triangle are each formed by a power of

11, and claimed that subsequent rows can also be generated by a power of 11. Literally, the claim is not true for

the 5 row and onward. His genius mind might have suggested a deep relation between binomial coeﬃcients

and a power of some integer that resembles the number 11 in some form. In this study, we propose and prove a

general formula to generate the values in any row of Pascal’s triangle from the digits of (1 0⋯0

Θzeros

1). It can be

shown that the numbers in the cells in row of Pascal’s triangle may be achieved from Θ +1 partitions of the

digits of the number (1 0⋯0

Θzeros

1), where Θis a non-negative integer. That is, we may generate the number in

the cells in a row of Pascal’s triangle from a power of 11, 101, 1001, or 10001 and so on. We brieﬂy discuss how

to determine the number of zeros Θin relation to , and then empirically show that the partition really gives

us binomial coeﬃcients for several values of . We provide a formula for Θand prove that the ( +1)

row of

Pascal’s triangle is simply Θ +1 partitions of the digits of (1 0⋯0

Θzeros

1)from the right.

1. Introduction

Algebra is a spacious part of the science of mathematics that provides the opportunity to express mathematical ideas precisely. In algebra, the

binomial expansion and Pascal’s triangle are considered important. Pascal’s triangle is an arrangement of the binomial coeﬃcients and one of the

most known integer models. Though it was named after the French scientist Blaise Pascal, it was studied in ancient India [1, 2], Persia [3, 4], China

[5], Germany, and Italy [6].

In reality, the deﬁnition of the triangle was made centuries ago. In 450 BC, an Indian mathematician named Pingala is said to have introduced

the deﬁnition of this triangle in a Sanskrit poetry book. Chinese mathematicians had the same idea and named the triangle as “Yang Hui’s triangle”.

Later, Persian mathematician Al-Karaji and Persian astronomer-poet Omar Khayyam named the triangle as the “Khayyam triangle”. It also has

multi-dimensional shapes. The three-dimensional shape is referred to as Pascal’s pyramid or Pascal’s tetrahedron, while the other general-shaped

ones are called Pascal’s simpliﬁcations.

Various studies have been conducted in many diﬀerent disciplines about Pascal’s triangle. For the construction of Pascal’s triangle, Sgroi [7]

stated that each line starts with 1 and ends with 1, and this series can be expanded with simple cross-joints. Jansson [8] developed three geometric

forms related to Pascal’s triangle and included examples of each form. Toschi [9] used various permutations to generate new forms of Pascal’s

triangles and generalized them. Duncan and Litwiller [10] addressed the reconstruction of Pascal’s triangle with the individuals. Here they collected

data on the opinions of individuals using qualitative methods, and determined the methods of constructing the Pascal’s triangle in diﬀerent ways

with the attained ﬁndings.

*Corresponding author.

E-mail address: robiul258@gmail.com (Md. Robiul Islam).

https://doi.org/10.1016/j.heliyon.2022.e11651

Received 29 December 2021; Received in revised form 6 July 2022; Accepted 8 November 2022

2405-8440/©2022 The Authors. Published by Elsevier Ltd. This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/).

Md. Shariful Islam, Md. Robiul Islam, Md. Shorif Hossan et al. Heliyon 8 (2022) e11651

Researchers worked on Pascal’s fascinating characteristics. Using the principle of permutation, Putz [11] designed Pascal Polytope and linked

it to the Fibonacci concept. Houghton [12] gave the concept of the relationship between the successive diﬀerential operations of a function and

Pascal’s triangle. With an application, he attempted to incorporate the idea of a diﬀerentiable function into Pascal’s triangle. The relationship

between Pascal’s triangle and the Tower of Hanoi has been elucidated by Andreas M. Hinz [13]. Finding diagonal sum [14], k-Fibonacci sequence,

recurrence relations [15], ﬁnding exponential ()[16] were a part of those to describe the work that is generated from Pascal’s triangle. Some

fascinating properties of Pascal’s triangle are available in [17, 18]. In 1956, Freund [19] elicited that the generalized Pascal’s triangles of th

order can be constructed from the generalized binomial coeﬃcients of order . Bankier [20] gave the Freud’s alternative proof. Kallós generalized

Pascal’s triangle from an algebraic point of view by diﬀerent bases. He tried to generalize Pascal’s triangle using the power of integers [21], powers

of base numbers [22] and their connections with prime numbers [23]. Kuhlmann tried to generate Pascal’s triangle using the T-triangle concept

[24].

The concept of using a power of 11 to generate rows of Pascal’s triangle was ﬁrst introduced by Sir Isaac Newton. He noticed the ﬁrst ﬁve rows

of Pascal’s triangle are formed by a power of 11 and claimed (without proof) that subsequent rows can also be generated by a power of eleven as

well [25]. Arnold et al. [26] showed if one assigns a place value to each of the individual terms in a certain row of the triangle, the pattern can be

seen again. Morton [27] noted the Pascal’s triangle property by the power of 11 for 10 base numeral system. Mueller [28] noted that one can get

the th power of 11 from the th row of the Pascal’s triangle with positional addition.

It is clearly concluded that above mentioned works did not express a full row of Pascal’s triangle from a power of 11, or from the digits of

(1 0⋯0

Θzeros

1), as Sir Isaac Newton hinted. This paper has worked on the extension from powers of 11 to powers of 101, 1001, 10001, … (1 0⋯0

Θzeros

1) and

proved a new general formula to generate any row of Pascal’s triangle.

2. Methods

The very basic deﬁnition of getting the number at any cell of a row of Pascal’s triangle is the summation of the numbers at the two adjacent cells

of the previous row. The rows of Pascal’s triangle are numbered starting from =0on the top and the cells in each row are numbered from =0 on

the left. For =0, there is only one cell with the value 1. As the successive rows are generated, the numbers in the right most and left most cells are

deﬁned to be 1.

1

11

121

13 31

14641

15101051

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

1 9 36 84 126 126 84 36 9 1

…………

Fig. 1. Pascal’s triangle.

The power of 11 technique is generating Pascal’s triangle by multiplying previous rows by 11 successively. The one digit partition of 111=11

gives us the numbers in the cells of the 1st row and 112= 121, 113= 1331 and 114= 14641 give 2nd, 3rd, and 4th rows respectively. Before ﬁnding the

general rule for subsequent rows, we ﬁrst elaborate on the concept of powers of 11. The reason behind getting Pascal’s triangle from the powers of

11 lies in the general rule of multiplication. What do we get from multiplication of a number by 11? Let be the number generated by concatenating

each of the digits in the cells of the th row of Pascal’s triangle from left to right.

2 row of Pascal’s triangle →121

×11

121

left shift of all digits by 1 place →1210

3 row of Pascal’s triangle →1331

Fig. 2. Multiplication of 2 row by 11.

Fig. 2shows that multiplication of 121 by 11 gives 3. That is 3=112.

4 row of Pascal’s triangle →14641

×11

14641

left shift of all digits by 1 place →146410

not 5 row of Pascal’s triangle →161051

Fig. 3. Multiplication of 4 row by 11.

Fig. 1indicates that 5should be 15101051, whereas from Fig. 3we get 161051. So, we can make a comment from Fig. 3that multiplication of 4

by 11 does not give 5.

2

Md. Shariful Islam, Md. Robiul Islam, Md. Shorif Hossan et al. Heliyon 8 (2022) e11651

Patently 115= 161051 and 116= 1771561, but the 5th and 6th row of Pascal’s triangle are

1 5 10 10 5 1

and

1 6 15 20 15 6 1

respectively. The above scheme fails for 115or 116. Why does the power of 11 technique fail here, and why does the power of 11 technique work for

the ﬁrst four rows? If the reader closely looks at the Pascal’s triangle, they will see that all of the cell values in the ﬁrst to fourth rows are one digit.

We get two-digit cell values for the ﬁrst time in the central cells of the ﬁfth row, which we think is a potential reason for the power of 11 technique

failing here. So for ﬁnding the 5th row onward, we need two (three, four, …) digits partitions of . The shifting of places in Fig. 2and Fig. 3implies

using a power of 1 0⋯0

Θzeros

1, for some Θ, might work. Now, we will endeavor to formulate a speciﬁc rule.

At ﬁrst, we attempt to generate the number for which two digit partitions give us the numbers in the cells of a row of the Pascal’s triangle. So

we extend the concept of power of 11 technique to the power of 101 technique and multiply 101 by itself to see the consequences. We can achieve

this by using the very basic rules of multiplication.

101

×101

101

zeros cause the left shift of all digits by 1 place →0000

left shift of all digits by 2 places →10100

2 digits representation of 2 row →10201

×101

10201

000000

left shift of all digits by 2 places →1020100

2 digits representation of 3 row →1030301

Fig. 4. Eﬀects of multiplying by 101.

Fig. 4displays the impact of multiplication by 101. The result of 101 × 101 is 10201 whereas 11 ×11 = 121. One digit partition of 121 produce

121but two digits partition of 10201 yields 10201which is identical to 01 02 01. The colored pairs of digits in each product are the summation of

two numbers in the adjacent cells of the previous row.

Now, 1015= 10510100501, from which we can construct 5th row of Pascal’s triangle by two digits partitions from the right.

1 05 10 10 05 01

Similarly from 1016= 1061520150601 and 1017= 107213535210701, we can easily construct the 6th and 7th row respectively.

1 06 15 20 15 06 01

and

1 07 21 35 35 21 07 01 .

Hence, two-digit partitions of 1015, 1016and 1017generate the numbers in the cells of the 5, 6 and 7 rows of Pascal’s triangle, respectively, due

to the insertion of one zero between 1and 1in 11. Sir Issac Newton might have meant this technique in his claim.

Can a conclusion be drawn for generating the numbers in cells of any row of Pascal’s triangle with the help of some extended power of 11 technique

such as 101? The 9th row of Pascal’s triangle is

1 9 36 84 126 126 84 36 9 1

Clearly, two digits partition from the right of the number 1019= 1093685272684360901 does not give the numbers in the cells of the 9th row because

the numbers in the central cells of this row contains three digits.

So the representation of three place values for each entry of Pascal’s triangle requires a new formula to be generated. The previous context

directed that multiplication of a number by 11 and 101 makes the left shift of all digits by one and two places, respectively. Therefore, three-digit

representation requires multiplication by 1001.

Fig. 5indicates the left shift of all digits by 3places when a number is multiplied by 1001.

By continuing the multiplication by 1001 in Fig. 4, we get

10019= 1009036084126126084036009001

from which one may form the 9 row of Pascal’s triangle by three digits partition of the number from the right.

1 009 036 084 126 126 084 036 009 001

Similarly, (1001)10 = 1010045120210252210120045010001,

1010045120210252210120045010001 ⟼1 010 045 120 210 252 210 120 045 010 001

3

Md. Shariful Islam, Md. Robiul Islam, Md. Shorif Hossan et al. Heliyon 8 (2022) e11651

1001

×1001

1001

00000

000000

left shift of all digits by 3 time →1001000

3 digits representation of 2 row →1002001

×1001

1002001

00000000

000000000

left shift of all digits by 3 time →1002001000

3 digits representation of 3 row →1003003001

Fig. 5. Eﬀects of multiplying by 1001.

the 10th row of the Pascal’s triangle.

From the above study, it may be concluded that the three-digit representation requires the left shift of all digits by three places, and requires

two zeros between 1and 1in 11, that is 1001. Why do we require three-digit representation for the 9 and 10 rows of Pascal’s triangle? Because

the central cells of 9th and 10th rows are of three digits. Similarly, we need two-digit representation for 5 to 8 rows since the central cell of these

rows are numbers of two digits. And, the ﬁrst four rows are satisﬁed by 11since the central cell of the ﬁrst four rows contains one digit only. So

for any given row, the number of digits in the representation for the number in a cell should be equal to the number of digits in the central cell(s)

of that row. Why the central cell value(s) should be taken into account in this situation may be questioned. The central cell value(s) matter here

because an observation of Pascal’s triangle is that for any row, the central values are the largest of any other cell values of that row. So, knowing the

number of digits in the central cell value(s) implies the required number of digits in the partitioning of the representation for that particular row of

Pascal’s triangle.

The above discussion compels us to generate a formula to ﬁnd the central values of any row of the Pascal’s triangle. For an odd number, say

=9, we get +1 =10 elements in 9 row and so the central value should be

10

2

=5

observation of that row, which is

9

5−1=9

4= 126. For

an even number, say =10, we get +1 =11elements and the central value should be 11

2 =6

observation, which is

10

6−1=10

5= 252.

By taking the ﬂoor value of

2, a formula for central value of row is

2.

But we never need a central element; rather it is necessary to know how many digits the central element has. Applying the property of the logarithmic

function, one can identify how many digits (or place values) the central element has without knowing the value of the cell. Therefore, the number

of digits in the central value is given by

log10

2

since log10()represents the number of digits of when ≠10, for ∈ℕ. For a central value with Θ −1 digits we require Θzeros between

1and 1in 11, such that (1 0⋯0

Θzeros

1). So, the required number of zeros between 1and 1in 11 can be obtained by taking the ﬂoor value of log10

2.

If Θrepresents the number of zeros between 1and 1in 11, then

Θ=log10

2

We now verify it for an odd number =9and an even number =10.

If, =9 then Θ =2, and if =10then Θ =2.

For both of the numbers, we need 2zeros between 1and 1in 11. So, to get the 9 and 10 rows we have to calculate 10019and 100110

respectively. Both of these cases have been discussed above, and are consistent with our formula for Θ.

It’s time to generate the formula to ﬁnd any row of Pascal’s triangle. We infer that the general formula for generating the row of Pascal’s

triangle is the Θ +1 digit partitioning of the digits of the number (1 0⋯0

Θzeros

1)from the right. For =15, we get Θ =3. So, we have to insert 3zeros

between 1and 1in 11 and the 15 row can be constructed by four-digit partitions of the digits of the number

1000115 = 1001501050455136530035005643564355005300313650455010500150001

from the right as shown below

1 0015 0105 0455 1365 3003 5005 6435 6435 5005 3003 1365 0455 0105 0015 0001

Notice the partitioning yields the 15th row of the Pascal’s triangle

1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1

Similarly, we may verify for =16, Θ =4and

(100001)16 = 100016001200056001820043680800811440128701144008008043680182000560001200001600001

4

Md. Shariful Islam, Md. Robiul Islam, Md. Shorif Hossan et al. Heliyon 8 (2022) e11651

This 16th row can also be veriﬁed from the existing Pascal’s triangle. The above formula can be used for a large . We now exemplify the 51 row

of Pascal’s triangle. Hence =51 gives Θ =14.

We have to put 14 zeros between 1and 1in 11, that is (1000000000000001)51.

(1000000000000001)51 =1000000000000051000000000001275000000000020825000000000249900000000002349060000000018009460000000115775100000

0006367630500000030423123500000127777118700000476260169700001587533899000004762601697000012927061749000031886752314200071745192706

9501477106908672502790090827492504845947226697507753515562716011445665830676015607726132740019679306863020022959191340190024795926

6474052247959266474052229591913401900196793068630200156077261327400114456658306760077535155627160048459472266975027900908274925014

7710690867250071745192706950031886752314200012927061749000004762601697000015875338990000004762601697000001277771187000000304231235

0000000636763050000000115775100000000018009460000000002349060000000000249900000000000020825000000000001275000000000000051000000000

000001

The desired 51st row can be obtained by partitioning each 15 digits from the right. For readers’ convenience, we marked each partition with

diﬀerent colors and showed that the above formula generates the 51st row of the Pascal’s triangle.

3. Results and discussion

Remark: In general the th partition of length of the digits of a positive integer is the left most digits of the number

mod 10×.

Now we give a proof of the power of 11 technique. To prove the main theorem, we prove some inequalities and lemmas.

For ∈ℕ, we have the following inequalities

Θ=log10

2

From the property of ﬂoor function, Θ ≤log10

2<Θ +1

⇒10Θ≤

2<10Θ+1 (1)

⇒

2<10Θ+1

Since both sides of the above inequality are integers, the diﬀerence between 10Θ+1 and

2is at least 1, therefore

⇒10Θ+1 −

2≥1

⇒10Θ+1 −1≥

2(2)

From inequality (1), we also have

10Θ≤

2

For , ∈ℕand 0 ≤ ≤, the maximum value of

occurs when =

2. Hence

2≥

,(3)

and notice that (1 0⋯0

Θzeros

1)=10Θ+1 +1

.

Lemma 1. If , ∈ℕand 0 ≤ ≤, then

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1<10(Θ+1).

Proof. By inequality (3), we have

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1

≤

210(−1)(Θ+1) +

210(−2)(Θ+1) +⋯+

2

=

210(−1)(Θ+1) +10

(−2)(Θ+1) +⋯+1

(4)

Since, 10(−1)(Θ+1) +10

(−2)(Θ+1) +⋯ +1is a geometric series of terms with common ratio 10(Θ+1),

−1

=0

10(Θ+1)=1+10

(Θ+1) +⋯+10

(−1)(Θ+1) =10(Θ+1) −1

10(Θ+1) −1 (5)

5

Md. Shariful Islam, Md. Robiul Islam, Md. Shorif Hossan et al. Heliyon 8 (2022) e11651

From inequality (4) and equation (5) we have,

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1≤

210(Θ+1) −1

10(Θ+1) −1 <

210(Θ+1)

10(Θ+1) −1 (6)

From inequality (2), we have

210(Θ+1)

10(Θ+1) −1 ≤(10(Θ+1) −1) 10(Θ+1)

10(Θ+1) −1 =10

(Θ+1) (7)

From inequalities (6) and (7), we have

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1<10(Θ+1).□

Proposition 1. If , ∈ℕand 0 ≤ ≤, then

10Θ+1 +1

mod 10(Θ+1) =

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1

Proof. Expanding

10Θ+1 +1

by binomial theorem, we have

10Θ+1 +1

mod 10(Θ+1)

=

=0

−10(Θ+1){−(−)} mod 10(Θ+1)

=

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1 mod10

(Θ+1)

by Lemma 1, we have

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1<10(Θ+1)

therefore,

10Θ+1 +1

mod 10(Θ+1) =

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1 □

Corollary 1. The integer

−(−1)10(−1)(Θ+1) +

−(−2)10(−2)(Θ+1) +⋯ +1has at most (Θ +1)signiﬁcant digits.

Proof. This follows directly from how

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1

is the remainder when

10(Θ+1) +1

mod 10(Θ+1).□

Corollary 2. The left most partition of length

(Θ+1

)from the right of the integer

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1 is

−(−1)

=

−1

.

Proof. By Corollary 1, the integer

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1

has at most (Θ +1)signiﬁcant digits, and similarly

−(−2)

10(−2)(Θ+1) +

−(−3)

10(−3)(Θ+1) +⋯+1

has at most ( − 1)(Θ +1) signiﬁcant digits. Since

−(−1)10(−1)(Θ+1) has ( − 1)(Θ +1)zeros to the right and it has at most (Θ +1)signiﬁcant digits,

the left most partition of length Θ +1of

−(−1)10(−1)(Θ+1) and

−(−1)10(−1)(Θ+1) +

−(−2)10(−2)(Θ+1) +⋯ +1are the same.

Since the left most partition of

−(−1)10(−1)(Θ+1) is

−(−1)

=

−1

,

the left most partition of

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1 is

−1

.□

6

Md. Shariful Islam, Md. Robiul Islam, Md. Shorif Hossan et al. Heliyon 8 (2022) e11651

Theorem 1. The th partition of (Θ +1) digits from the right of the integer (1 0⋯0

Θzeros

1)is the binomial coeﬃcient

−1, where Θ =log10

2.

Proof. The th partition of (Θ +1)digits from the right of the integer (1 0⋯0

Θzeros

1)=10Θ+1 +1

is the left most partition of (1 0⋯0

Θzeros

1)mod 10×(Θ+1).

From Proposition 1,

(1 0 ⋯0

Θzeros

1)mod 10×(Θ+1) =

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1

Again from Corollary 2, the left most (Θ +1) many digits of

−(−1)

10(−1)(Θ+1) +

−(−2)

10(−2)(Θ+1) +⋯+1 is

−1

.□

Hence, the Θ +1 digits partition from the right of the digits of the integer (1 0⋯0

Θzeros

1)generates all the binomial coeﬃcients or the numbers in

the cells of the th row of the Pascal’s triangle.

4. Conclusion

Sir Isaac Newton hinted that binomial coeﬃcients in the th row of the Pascal’s triangle may be achieved from partitioning the digits in the th

power of some number that contains 11 in some form [25]. It has been shown earlier that the weighted sum of the values in the th row of the

Pascal’s triangle is (11)[26]. We have proved that

(Θ+1

)digit partitions of the digits of (1 0⋯0

Θzeros

1)from the right give the values of the cells in

the th row of the Pascal’s triangle, and provided an explicit formula for the value of Θas a function only of n.

Declarations

Author contribution statement

Md. Shariful Islam: Performed the experiments; Contributed reagents, materials, analysis tools or data.

Md. Robiul Islam: Conceived and designed the experiments; Performed the experiments; Analyzed and interpreted the data; Wrote the paper.

Md. Shorif Hossan: Conceived and designed the experiments; Analyzed and interpreted the data; Contributed reagents, materials, analysis tools

or data; Wrote the paper.

Md. Hasan Kibria: Performed the experiments.

Funding statement

This research did not receive any speciﬁc grant from funding agencies in the public, commercial, or not-for-proﬁt sectors.

Data availability statement

No data was used for the research described in the article.

Declaration of interests statement

The authors declare no conﬂict of interest.

Additional information

No additional information is available for this paper.

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