New presences of $\pi$ and $e$ in Pascal's triangle

Abstract

The following work shows new connections between the constants $\pi$ and $e$ with Pascal's triangle and the Lucas triangle, established via Fibonacci polynomials and similar means. Furthermore, relations between the two famous constants and the rows of Pascal's triangle and the Lucas triangle are conjectured, together with some other important related identities.

Full text

New presences of πand ein Pascal’s triangle
Mauricio Guevara Valerio
University of Costa Rica
mauricio.guevaravalerio@ucr.ac.cr
Abstract: The following work shows new connections between the constants πand e
with Pascal’s triangle and the Lucas triangle, established via Fibonacci polynomials and
similar means. © 2023 The Author(s)
1. Background
Discoveries of eand πin Pascal’s triangle are relatively recent in the history of mathematics. In the case of e,
Harlan J. Brothers found the following remarkable relation in 2012 [1]:
lim
n→∞
sn+1
sn
sn
sn−1
=e
Where snis the product of all the entries in the nth row of Pascal’s triangle.
It is less known, however, that the great popularizer of mathematics Martin Gardner had already mentioned a
relation between eand the triangle through the Fibonacci numbers [2], namely the following (although as he
stated it the claim is false):
e=1+1+2
2! +3
3! +5
4! +8
5! +13
6! +21
7! +34
8! +55
9! +···
1+0+1
2! +1
3! +2
4! +3
5! +5
6! +8
7! +13
8! +21
9! +···
A slight alteration of the signs is all it takes to get the true identity:
e=1+1+2
2! +3
3! +5
4! +8
5! +13
6! +21
7! +34
8! +55
9! +···
1−0+1
2! −1
3! +2
4! −3
5! +5
6! −8
7! +13
8! −21
9! +···
Gardner does not mention who discovered this curious identity, nor how to prove it. He — or rather his character
“O’Shea” — says only that it was found “on the Web”.
Although this identity is very remarkable on its own, since it shows the connection between eand the Fibonacci
numbers, it turns out there is a similar and more interesting relationship between eand the Fibonacci polynomials,
through which a relationship between e,πand Pascal’s triangle can be established . The structure of this work
is as follows: in section 2 it will be shown how the identity mentioned by Gardner can be proved after the signs
are fixed, using a more general formula relating eand Fibonacci numbers, as well as other Lucas sequences. In
section 3 a similar relation between eand the Fibonacci Polynomials is established. In section 4 the connection
with πis established. Finally, in section 5 some other important related identities are demonstrated, together with
a conjecture related to the other shallow diagonals of Pascal’s triangle .
2. Identities relating e to Fibonacci numbers and other Lucas sequences
Theorem 1:
e=
∞
∑
k=0
Fk+1+xFk−1
k!
∞
∑
k=0
(−1)k·Fk−1+xFk+1
k!

Proof: From the Euler - Binet formula:
Fn=φn−(1−φ)n
√5
And the definition of exas the sum of an infinite series:
ex=
∞
∑
k=0
xk
k!
It follows that:
∞
∑
k=0
Fk+1+xFk−1
k!
∞
∑
k=0
(−1)k·Fk−1+xFk+1
k!
=
∞
∑
k=0
φk+1−(1−φ)k+1+xφk−1−x(1−φ)k−1
√5·k!
∞
∑
k=0
(−1)k·φk−1−(1−φ)k−1+xφk+1−x(1−φ)k+1
√5·k!
It can be verified [3] that this last expression is equal to:
=
2e1−φ(x√5e√5−xe√5+x√5+x+√5e√5+e√5+√5−1)
4√5
2e−φ(x√5e√5−xe√5+x√5+x+√5e√5+e√5+√5−1)
4√5
From here the identity follows almost immediately:
=2e1−φ
2e−φ=e1−φ
e−φ=e1−φ+φ=e
And so we have proved our first Theorem. To get the identity Gardner mentions, we just have to set x=0 in
Theorem 1:
e=
∞
∑
k=0
Fk+1
k!
∞
∑
k=0
(−1)k·Fk−1
k!
If we set x = 1, we get a relation between eand the Lucas numbers:
e=
2
0! +1
1! +3
2! +4
3! +7
4! +11
5! +18
6! +29
7! +47
8! +76
9! +···
2
0! −1
1! +3
2! −4
3! +7
4! −11
5! +18
6! −29
7! +47
8! −76
9! +···
This identity can also be proved through a different formula, as may be seen in the final section. It is also worth
mentioning that Theorem 1 gives relations between e and Lucas sequences that also have the same recurrence
relation and link to the golden ratio, when x has positive integer values. For instance, setting x=2 gives the
following identity:
e=
3
0! +1
1! +4
2! +5
3! +9
4! +14
5! +23
6! +37
7! +60
8! +97
9! +···
3
0! −2
1! +5
2! −7
3! +12
4! −19
5! +31
6! −50
7! +81
8! −131
9! +···
It may also be worth noting that by writing down the terms with negative indices of the sequence 3,1,4,5,9,···
to the left, we can observe the following pattern:
··· ,−19,12,−7,5,−2,3,1,4,5,9,···
This pattern seems to hold for the other Lucas sequences as well.

3. The connection between eand Pascal’s triangle via Fibonacci polynomials
To prove the identity that relates Euler’s number with Pascal’s triangle through Fibonacci polynomials, we will
use the closed form formula for the Fibonacci polynomials [4] instead of Binet’s formula:
Fk(x) = (x+√x2+4)k−(x−√x2+4)k
2k√x2+4
Theorem 2:
ex=
∞
∑
k=1
Fk(x)
k!
∞
∑
k=1
(−1)k+1·Fk(x)
k!
Proof: From the closed form formula for the Fibonacci polynomials, it follows that:
∞
∑
k=1
Fk(x)
k!
∞
∑
k=1
(−1)k+1·Fk(x)
k!
=
∞
∑
k=1
(x+√x2+4)k
−(x−√x2+4)k
2k√x2+4
k!
∞
∑
k=1
(−1)k+1
(x+√x2+4)k
−(x−√x2+4)k
2k√x2+4
k!
=
1
√x2+4·
∞
∑
k=1
x+√x2+4
2k
−x−√x2+4
2k
k!
1
√x2+4·
∞
∑
k=1
(−1)k+1x+√x2+4
2k
−x−√x2+4
2k
k!
From the definition of exas an infinite series, it follows that:
=ex+√x2+4
2−ex−√x2+4
2
∞
∑
k=1
−−x+√x2+4
2k
−−−x−√x2+4
2k
k!
=ex+√x2+4
2−ex−√x2+4
2
∞
∑
k=1
−−x+√x2+4
2k
+−x−√x2+4
2k
k!
=ex+√x2+4
2−ex−√x2+4
2
e−x−√x2+4
2−e−x+√x2+4
2
=ex+√x2+4
2−ex−√x2+4
2
e−x+√x2+4
2−e−x−√x2+4
2
Now we just have to simplify a bit:
=e√x2+4−1·ex−√x2+4
2
e√x2+4−1·e−x−√x2+4
2
=ex−√x2+4
2
e−x−√x2+4
2
=ex−√x2+4
2·ex+√x2+4
2=e2x
2=ex
Thus completing the proof. To get to an identity very similar to the one mentioned by Gardner, we just have to set
x=1 in Theorem 2:
e=1+1
2! +2
3! +3
4! +5
5! +8
6! +13
7! +21
8! +34
9! +···
1−1
2! +2
3! −3
4! +5
5! −8
6! +13
7! −21
8! +34
9! −···
And it is known that the Fibonacci polynomials can be found in the shallow diagonals of Pascal’s triangle, if we
interpret its numbers as coefficients:

Fig. 1. Fibonacci polynomials in Pascal’s triangle
4. The connection between π,eand Pascal’s triangle via Fibonacci polynomials
Theorem 3:
e−1
e
2=π2
3! −π4−3π2
5! +π6−5π4+6π2
7! −π8−7π6+15π4−10π2
9! +···
Proof: From Euler’s identity:
eiπ+1=0→eiπ=−1
Setting x=iπin Theorem 2 yields:
eiπ=1+iπ
2! +(iπ)2+1
3! +(iπ)3+2iπ
4! +(iπ)4+3(iπ)2+1
5! +(iπ)5+4(iπ)3+3iπ
6! +···
1−iπ
2! +(iπ)2+1
3! −(iπ)3+2iπ
4! +(iπ)4+3(iπ)2+1
5! −(iπ)5+4(iπ)3+3iπ
6! +···
Combining the above with Euler’s identity gives the following identity:
−1=1+iπ
2! +(iπ)2+1
3! +(iπ)3+2iπ
4! +(iπ)4+3(iπ)2+1
5! +(iπ)5+4(iπ)3+3iπ
6! +···
1−iπ
2! +(iπ)2+1
3! −(iπ)3+2iπ
4! +(iπ)4+3(iπ)2+1
5! −(iπ)5+4(iπ)3+3iπ
6! +···
→ −1+iπ
2! −(iπ)2+1
3! +(iπ)3+2iπ
4! −(iπ)4+3(iπ)2+1
5! +(iπ)5+4(iπ)3+3iπ
6! −··· =
1+iπ
2! +(iπ)2+1
3! +(iπ)3+2iπ
4! +(iπ)4+3(iπ)2+1
5! +(iπ)5+4(iπ)3+3iπ
6! +···
Now all the terms whose denominator is an even factorial get cancelled out:
→ −1−(iπ)2+1
3! −(iπ)4+3(iπ)2+1
5! −··· =1+(iπ)2+1
3! +(iπ)4+3(iπ)2+1
5! +···
Since we have an identity of the form −x=x, that means both sides of the equation must equal 0. We can
therefore take the right hand side:
0=1+(iπ)2+1
3! +(iπ)4+3(iπ)2+1
5! +···=1−π2−1
3! +π4−3π2+1
5! −π6−5π4+6π2−1
7! +π8−7π6+15π4−10π2+1
9! −···
And use the Taylor series expansion for sinh(1):
sinh(1) = e−1
e
2=1+1
3! +1
5! +1
7! +1
9! +··· =1−(−1
3! ) + 1
5! −(−1
7! ) + 1
9! −···
By combining the two previous identities and simplifying, we get:
0=e−1
e
2−π2
3! +π4−3π2
5! −π6−5π4+6π2
7! +π8−7π6+15π4−10π2
9! −···
And finally we get the beautiful identity that relates e,πand Pascal’s triangle through Fibonacci polynomials,
completing the proof of Theorem 3:

Fig. 2. Extension of Pascal’s triangle to negative numbers: the “rotated” Pascal triangle
Fig. 3. Pattern of the coefficients for the powers of πin Theorem 3
e−1
e
2=π2
3! −π4−3π2
5! +π6−5π4+6π2
7! −π8−7π6+15π4−10π2
9! +···
The pattern in the coefficients of the powers of πcan be found in the “rotated” Pascal triangle, following the lines
in figure 3.
A possible way of proving Theorem 3 without using Euler’s identity is the following:
Conjecture 1.
1
1! =π2
3! −π4
5! +π6
7! −π8
9! +···
1
3! =3π2
5! −5π4
7! +7π6
9! −9π8
11! +···
1
5! =6π2
7! −15π4
9! +28π6
11! −45π8
13! +···
1
7! =10π2
9! −35π4
11! +84π6
13! −165π8
15! +···
And so on ad infinitum. Using Wolfram Alpha to verify those identities, it can be seen that this pattern seems to
hold. If this is true, then we can add the terms using a method similar to the proof that the set of all rational
numbers and the set of all natural numbers are equipotent:
Fig. 4. Zigzag proof of Theorem 3

A similar argument shows that:
sin(1) = π2
3! −π4+3π2
5! +π6+5π4+6π2
7! −π8+7π6+15π4+10π2
9! +···
5. Other related identities
In this section I will mention some other important identities that follow from or are similar to ones that have
been already mentioned.
A connection between e and πestablished via Chebyshev polynomials of the second kind can be deduced setting
x=2πiin Theorem 2:
e−1
e
2=8π2
4! −32π4−32π2
6! +128π6−192π4+80π2
8! −512π8−1024π6+672π4−160π2
10! +···
The proof of this identity is very analogous to the proof of Theorem 3. The only significant difference is that
since e2πi=1, the terms that get canceled out are the ones whose denominator is the factorial of an odd number.
It is also worth noting that many other interesting identities can be obtained through a similar connection between
eand the Lucas polynomials, through which a connection between e,πand the Lucas triangle can be established:
Theorem 4.
ex=
∞
∑
k=0
Lk(x)
k!
∞
∑
k=0
(−1)k·Lk(x)
k!
Proof: From the closed form formula for the Lucas polynomials [6]:
Lk(x) = 1
2k
·hx+px2+4k+x−px2+4ki
And the definition of exas an infinite series:
ex=
∞
∑
k=0
xk
k!
It follows that:
∞
∑
k=0
Lk(x)
k!
∞
∑
k=0
(−1)k·Lk(x)
k!
=
∞
∑
k=01
2khx+√x2+4k
+x−√x2+4ki
k!
∞
∑
k=0
(−1)k·1
2khx+√x2+4k
+x−√x2+4ki
k!
=
∞
∑
k=0x+√x2+4
2k
+x−√x2+4
2k
k!
∞
∑
k=0−x+√x2+4
2k
+−x−√x2+4
2k
k!
=
∞
∑
k=0x+√x2+4
2k
+x−√x2+4
2k
k!
∞
∑
k=0−x−√x2+4
2k
+−x+√x2+4
2k
k!
=ex+√x2+4
2+ex−√x2+4
2
e−x−√x2+4
2+e−x+√x2+4
2

=
ex−√x2+4
2e2√x2+4
2+1
e−x−√x2+4
2e2√x2+4
2+1
=ex−√x2+4
2
e−x−√x2+4
2
=ex−√x2+4
2·ex+√x2+4
2=e2x
2=ex
As was to be proved.
Setting x=1 gives again the identity for ein terms of the Lucas numbers:
e=
2
0! +1
1! +3
2! +4
3! +7
4! +11
5! +18
6! +29
7! +47
8! +76
9! +···
2
0! −1
1! +3
2! −4
3! +7
4! −11
5! +18
6! −29
7! +47
8! −76
9! +···
Other values of x also give interesting identities, and using the fact that Lucas numbers are powers of φrounded
to the nearest whole number, we can see them in this form (and similar identities can be obtained for the other
metallic means):
e4=
2
0! +[φ3]
1! +[φ6]
2! +[φ9]
3! +[φ12]
4! +[φ15]
5! +[φ18]
6! +···
2
0! −[φ3]
1! +[φ6]
2! −[φ9]
3! +[φ12]
4! −[φ15]
5! +[φ18]
6! −···
e11 =
2
0! +[φ5]
1! +[φ10]
2! +[φ15]
3! +[φ20]
4! +[φ25]
5! +[φ30]
6! +···
2
0! −[φ5]
1! +[φ10]
2! −[φ15]
3! +[φ20]
4! −[φ25]
5! +[φ30]
6! −···
e29 =
2
0! +[φ7]
1! +[φ14]
2! +[φ21]
3! +[φ28]
4! +[φ35]
5! +[φ42]
6! +···
2
0! −[φ7]
1! +[φ14]
2! −[φ21]
3! +[φ28]
4! −[φ35]
5! +[φ42]
6! −···
e76 =
2
0! +[φ9]
1! +[φ18]
2! +[φ27]
3! +[φ36]
4! +[φ45]
5! +[φ54]
6! +···
2
0! −[φ9]
1! +[φ18]
2! −[φ27]
3! +[φ36]
4! −[φ45]
5! +[φ54]
6! −···
And so on.
Setting x=iπin Theorem 4 gives the following identity after some straightforward simplification:
e+e−1=π2
2! −π4−4π2
4! +π6−6π4+9π2
6! −π8−8π6+20π4−16π2
8! +···
Fig. 5. Lucas polynomials in the shallow diagonals of the Lucas triangle
Here are a couple of questions that might be worth pondering: what about the (1,3) Pascal triangle? Can one find
identities for eand πanalogous to those of Pascal’s triangle and the Lucas triangle? And for (1,n)Pascal
triangles with n>3? It turns out one can (sort of), and it follows from the following generalization:
xe−e−1
2=yπ2−yπ4−(y+x)π2
3! +yπ6−(3y+x)π6+ (y+3x)π2
5! −yπ8−(5y+x)π6+ (6y+5x)π4−(y+6x)π2
7! +
yπ10 −(7y+x)π8+ (15y+7x)π6−(10y+15x)π4+ (y+10x)π2
9! −···
This seems to hold for all xand y. The coefficients follow the odd indexed shallow diagonals of this generalized
Pascal triangle:

Fig. 6. Generalized Pascal triangle
If we set y=1, we get a general relation with Pascal-type triangles of the form (1,x)
Moving on, a more general relation between eand the Lucas polynomials can be proved:
Theorem 5:
eLn(x)=
∞
∑
k=0
Lnk(x)
k!
∞
∑
k=0
(−1)kLnk(x)
k!
Proof:
∞
∑
k=0
Lnk(x)
k!
∞
∑
k=0
(−1)kLnk(x)
k!
=
∞
∑
k=0hx+√x2+4
2nik
k!+
∞
∑
k=0hx−√x2+4
2nik
k!
∞
∑
k=0h−x+√x2+4
2nik
k!+
∞
∑
k=0h−x−√x2+4
2nik
k!
=ex+√x2+4
2n
+ex−√x2+4
2n
1
ex+√x2+4
2n+1
ex−√x2+4
2n
=ex+√x2+4
2n
+ex−√x2+4
2n
ex+√x2+4
2n
+ex−√x2+4
2n
ex+√x2+4
2n
+x−√x2+4
2n
=ex+√x2+4
2n
+x−√x2+4
2n
=eLn(x)
As was to be proved. We can use this theorem to “fill in the gaps” in the series of identities for eand φ:
e3=
2
0! +[φ2]
1! +[φ4]
2! +[φ6]
3! +[φ8]
4! +[φ10]
5! +[φ12]
6! +···
2
0! −[φ2]
1! +[φ4]
2! −[φ6]
3! +[φ8]
4! −[φ10]
5! +[φ12]
6! −···

e7=
2
0! +[φ4]
1! +[φ8]
2! +[φ12]
3! +[φ16]
4! +[φ20]
5! +[φ24]
6! +···
2
0! −[φ4]
1! +[φ8]
2! −[φ12]
3! +[φ16]
4! −[φ20]
5! +[φ24]
6! −···
e18 =
2
0! +[φ6]
1! +[φ12]
2! +[φ18]
3! +[φ24]
4! +[φ30]
5! +[φ36]
6! +···
2
0! −[φ6]
1! +[φ12]
2! −[φ18]
3! +[φ24]
4! −[φ30]
5! +[φ36]
6! −···
e47 =
2
0! +[φ8]
1! +[φ16]
2! +[φ24]
3! +[φ32]
4! +[φ40]
5! +[φ48]
6! +···
2
0! −[φ8]
1! +[φ16]
2! −[φ24]
3! +[φ32]
4! −[φ40]
5! +[φ48]
6! −···
And so on.
Finally, here’s a surprising relation between the other shallow diagonals of Pascal’s triangle and its normal
diagonals:
Conjecture 2.
π
2! −π3−2π
4! +π5−4π3+3π
6! −π7−6π5+10π3−4π
8! +···=2 1
0!
π+
2
0!
π3+
6
0! −π2
2!
π5+
20
0! −4π2
2!
π7+
70
0! −15π2
2! +π4
4!
π9+···!
=
2sin√π2−4
2
√π2−4
Fig. 7. LHS in Conjecture 7

Fig. 8. RHS in Conjecture 7
If we add up the highlighted terms in the normal diagonals, we seem to get the sequence A014300 in the OEIS:
the convolution of the central binomial coefficients and the Fine numbers[7].

REFERENCES
[1] Harlan J. Brothers, Math bite: finding e in Pascal’s triangle, Math. Mag. 85 (1) (2012) 51.
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[3] Wolfram Alpha LLC. 2009. Wolfram—Alpha.
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