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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A359760 Triangle read by rows. The Kummer triangle, the coefficients of the Kummer polynomials. K(n, k) = binomial(n, k) * oddfactorial(k/2) if k is even, otherwise 0, where oddfactorial(z) := (2*z)!/(2^z*z!).

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 1, 0, 3, 0, 1, 0, 6, 0, 3, 1, 0, 10, 0, 15, 0, 1, 0, 15, 0, 45, 0, 15, 1, 0, 21, 0, 105, 0, 105, 0, 1, 0, 28, 0, 210, 0, 420, 0, 105, 1, 0, 36, 0, 378, 0, 1260, 0, 945, 0, 1, 0, 45, 0, 630, 0, 3150, 0, 4725, 0, 945, 1, 0, 55, 0, 990, 0, 6930, 0, 17325, 0, 10395, 0
Offset: 0

Views

Author

Peter Luschny, Jan 13 2023

Keywords

Comments

The Kummer numbers K(n, k) are a refinement of the oddfactorial numbers (A001147) in the sense that they are the coefficients of polynomials K(n, x) = Sum_{n..k} K(n, k) * x^k that take the value oddfactorial(n) at x = 1. The coefficients of x^n are the aerated oddfactorial numbers A123023.
These numbers appear in many different versions (see the crossrefs). They are the coefficients of the Chebyshev-Hermite polynomials in signed form when ordered in decreasing powers. Our exposition is based on the seminal paper by Kummer, which preceded the work of Chebyshev and Hermite for more than 20 years. They are also referred to as Bessel numbers of the second kind (Mansour et al.) when the odd powers are omitted.

Examples

			Triangle K(n, k) starts:
 [0] 1;
 [1] 1, 0;
 [2] 1, 0,  1;
 [3] 1, 0,  3, 0;
 [4] 1, 0,  6, 0,   3;
 [5] 1, 0, 10, 0,  15, 0;
 [6] 1, 0, 15, 0,  45, 0,   15;
 [7] 1, 0, 21, 0, 105, 0,  105, 0;
 [8] 1, 0, 28, 0, 210, 0,  420, 0, 105;
 [9] 1, 0, 36, 0, 378, 0, 1260, 0, 945, 0;

References

  • John Riordan, Introduction to Combinatorial Analysis, Dover (2002), pp. 85-86.

Links

Crossrefs

Variants: Signed version: A073278. Other variants are the irregular triangle A100861 with zeros deleted, A066325 and A099174 with reversed rows, A111924, A144299, A104556.
Cf. A000085, A001147, A123023, A047974, A115329, A002522, A056107, A359739,
A001879, A001464, A005425, A344501, A123022, A359761.

Programs

  • Maple
    oddfactorial := proc(z) (2*z)! / (2^z*z!) end:
K := (n, k) -> ifelse(irem(k, 2) = 1, 0, binomial(n, k) * oddfactorial(k/2)):
seq(seq(K(n, k), k = 0..n), n = 0..11);
# Alternative, as coefficients of polynomials:
p := (n, x) -> 2^(n/2)*(-1/x^2)^(-n/2)*KummerU(-n/2, 1/2, -1/(2*x^2)):
seq(print(seq(coeff(simplify(p(n, x)), x, k), k = 0..n)), n = 0 ..9);
# Using the exponential generating function:
egf := exp(x + (t*x)^2 / 2): ser := series(egf, x, 12):
seq(print(seq(coeff(n! * coeff(ser, x, n), t, k), k = 0..n)), n = 0..9);
  • Mathematica
    K[n_, k_] := K[n, k] = Which[OddQ[k], 0, k == 0, 1, n == k, K[n - 1, n - 2], True, K[n - 1, k] n/(n - k)];
Table[K[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 25 2023 *)
  • Python
    from functools import cache
@cache
def K(n: int, k: int) -> int:
    if k %  2: return 0
    if n <  3: return 1
    if n == k: return K(n - 1, n - 2)
    return (K(n - 1, k) * n) // (n - k)
for n in range(10): print([K(n, k) for k in range(n + 1)])

Formula

Let p(n, x) = 2^(n/2)*(-1/x^2)^(-n/2)*KummerU(-n/2, 1/2, -1/(2*x^2)).
p(n, 1) = A000085(n); p(n, sqrt(2)) = A047974(n); p(n, 2) = A115329(n);
p(2, n) = A002522(n) (n >= 1); p(3, n) = A056107(n) (n >= 1);
p(n, n) = A359739(n) (n >= 1); 2^n*p(n, 1/2) = A005425(n).
K(n, k) = [x^k] p(n, x).
K(n, k) = [t^k] (n! * [x^n] exp(x + (t*x)^2 / 2)).
K(n, n) = A123023(n).
K(n, n-1) = A123023(n + 1).
K(2*n, 2*n) = A001147(n).
K(4*n, 2*n) = A359761, the central terms without zeros.
K(2*n+2, 2*n) = A001879.
Sum_{k=0..n} (-1)^n * i^k * K(n, k) = A001464(n), ((the number of even involutions) - (the number of odd involutions) in the symmetric group S_n (Robert Israel)).
Sum_{k=0..n} Sum_{j=0..k} K(n, j) = A000085(n + 1).
For a recursion see the Python program.

A268533 Pascal's difference pyramid read first by blocks and then by rows: T(n,k,m) = 1/(m!) * (d/dx)^m((1-x)^k*(1+x)^(n-k))|_{x=0}.

Original entry on oeis.org

1, 1, 1, 1, -1, 1, 2, 1, 1, 0, -1, 1, -2, 1, 1, 3, 3, 1, 1, 1, -1, -1, 1, -1, -1, 1, 1, -3, 3, -1, 1, 4, 6, 4, 1, 1, 2, 0, -2, -1, 1, 0, -2, 0, 1, 1, -2, 0, 2, -1, 1, -4, 6, -4, 1, 1, 5, 10, 10, 5, 1, 1, 3, 2, -2, -3, -1, 1, 1, -2, -2, 1, 1, 1, -1, -2, 2, 1, -1, 1, -3, 2, 2, -3, 1, 1, -5, 10, -10, 5, -1
Offset: 0

Views

Author

Bradley Klee, Feb 22 2016

Keywords

Comments

T(n,k,m) is a pyramidal stack of (n+1) X (n+1)-dimensional matrices, or an infinite-dimensional matrix in block-diagonal form (see examples).
Define triangular slices T_x(i,j) = T(2x+i,x,x+j) with i in {0,1,...} and j in {0,1,... i}. T_0 is Pascal's triangle, and it appears that T_{x} is a triangle of first differences T_{x}(i,j) = T_{x-1}(i+1,j+1)-T_{x-1}(i+1,j) (cf. A007318, A214292).
The so-called "quantum Pascal's pyramid", denoted QT(n,k,m), is obtained from Pascal's pyramid by a complexification of matrix elements: QT(n,k,m) = (-1)^(3m/2) T(n,k,m). QT(n,k,m) effects a Hermite-Cartesian (cf. A066325) to Laguerre-polar change of coordinates (see examples).
Row reversal is complex conjugation: QT(n,n-k,m) = QT(n,k,m)*.
To construct the "normalized quantum Pascal's pyramid", NQT(n,k,m), we need normalization numerators, NumT(n,k,m) as in A269301, and denominators, DenT(n,k,m) as in A269302; then, NQT(n,k,m) = sqrt(NumT(n,k,m) / DenT(n,k,m)) QT(n,k,m). In the context of physics NQT(n,k,m) acting as matrix conjugation effects a cyclic permutation of the infinite-dimensional generators of rotation, so NQT(n,k,m) is essentially equivalent to an infinite-dimensional rotation with (z,y,z) Euler angles (0,Pi/2,Pi/2) (Harter, Klee, see examples).
Normalization or no, Pascal's pyramid also arises in laser optics (Allen et al.) as the paraxial wave equation often admits a useful analogy to the Schrödinger equation for the two-dimensional isotropic quantum harmonic oscillator.

Examples

			First few blocks:
1
.  1,  1
.  1, -1
. . . . .  1,  2,  1
. . . . .  1,  0, -1
. . . . .  1, -2,  1
. . . . . . . . . . .  1,  3,  3,  1
Second triangle . . .  1,  1, -1, -1
slice, T_1: . . . . .  1, -1, -1,  1
0 . . . . . . . . . .  1, -3,  3, -1
1  -1 . . . . . . . . . . . . . . . .  1,  4,  6,  4,  1
2   0  -2 . . . . . . . . . . . . . .  1,  2,  0, -2, -1
3,  2, -2, -3 . . . . . . . . . . . .  1,  0, -2,  0,  1
4,  5,  0, -5, -4 . . . . . . . . . .  1, -2,  0,  2, -1
5,  9,  5, -5, -9, -5 . . . . . . . .  1, -4,  6, -4,  1
n=2 Cartesian/Polar coordinate change using quantum Pascal's pyramid:
| 1  -2 i  -1 |   | y^2 - 1 |    | - (r exp[ I \phi])^2 |
| 1   0     1 | * |   x*y   | =  |      r^2  -  2       |
| 1   2 i  -1 |   | x^2 - 1 |    | - (r exp[-I \phi])^2 |
When: x = r cos[\phi], y= r sin[\phi].
Permutation of Pauli Matrices, \sigma_i, using normalized quantum Pascal's pyramid:
                  | 1  -i |
R = (1/sqrt[2]) * | 1   i |
Then, R * \sigma_j * R^{\dagger} = \sigma_{pi(j)},
where pi(j) is a cyclic permutation: { 1 -> 2, 2 -> 3, 3 -> 1 }.

References

  • L. Allen, S. M. Barnett, and M. J. Padgett, Optical angular momentum, Institute of Physics Publishing, Bristol, 2003.

Links

Crossrefs

Cf. A007318, A214292, A269301, A269302.

Programs

  • Mathematica
    PascalsPyramid[Block_] := Outer[Simplify[Function[{n, k, m},1/(m!)(D[(1 - x)^k*(1 + x)^(n - k), {x, m}] /. x -> 0)][Block, #1, #2]] &, Range[0, Block], Range[0, Block]]; PascalsPyramid /@ Range[0, 10]

Formula

T(n,k,m) = (1/(m!)) * (d/dx)^m((1-x)^k*(1+x)^(n-k))|_{x=0}.

A331816 Irregular triangle (read by rows) of coefficients T(n,k) of polynomials p(n,x) = Sum_{k=0..2*n} T(n,k) * x^k = (-1)^n * e^(x^3/3) * (((d/dx)^n) e^(-x^3/3)) for n >= 0 and 0 <= k <= 2*n.

Original entry on oeis.org

1, 0, 0, 1, 0, -2, 0, 0, 1, 2, 0, 0, -6, 0, 0, 1, 0, 0, 20, 0, 0, -12, 0, 0, 1, 0, -40, 0, 0, 80, 0, 0, -20, 0, 0, 1, 40, 0, 0, -360, 0, 0, 220, 0, 0, -30, 0, 0, 1, 0, 0, 1120, 0, 0, -1680, 0, 0, 490, 0, 0, -42, 0, 0, 1, 0, -2240, 0, 0, 9520, 0, 0, -5600, 0, 0, 952, 0, 0, -56, 0, 0, 1
Offset: 0

Views

Author

Werner Schulte, Jan 27 2020

Keywords

Comments

Let r(s;n,x) = Sum_{k=0..s*n} A(s;n,k)*x^k = (-1)^n * e^(x^(s+1)/(s+1)) * (((d/dx)^n) e^(-x^(s+1)/(s+1))) for n >= 0 and x complex and some fixed integer s >= 1. Special cases: A(1;n,k) = A066325(n,k) and A(2;n,k) is this triangle. Formula: A(s;n,k) = (Sum_{i=0..floor(k/(s+1))} (-1)^i * binomial((n+k) /(s+1),i) * binomial(n+k-(s+1)*i,n)) * (-1)^(n-(n+k)/(s+1)) * (n!) / ((s+1)^((n+k)/(s+1)) * (((n+k)/(s+1))!)) if (n+k) mod (s+1) = 0 else 0 with n >= 0 and 0 <= k <= s*n.
Recurrence: (1) A(s;n,k) = A(s;n-1,k-s) - (k+1) * A(s;n-1,k+1),
(2) r(s;n,x) = x^s * r(s;n-1,x) - ((d/dx) r(s;n-1,x)) for n > 0 with initial values A(s;0,0) = 1 = r(s;0,x) and A(s;n,k) = 0 if k < 0 or k > s*n or (n+k) mod (s+1) > 0;
E.g.f.: Sum_{n>=0} r(s;n,x)*t^n/(n!) = e^((x^(s+1)-(x-t)^(s+1))/(s+1)).
This generalization is result of a long and intensive discussion with Wolfdieter Lang. For more information see A091752.

Examples

			The irregular triangle T(n,k) starts:
n\k:  0     1    2    3    4     5   6     7   8   9  10   . . .      16
========================================================================
0  :  1
1  :  0     0    1
2  :  0    -2    0    0    1
3  :  2     0    0   -6    0     0   1
4  :  0     0   20    0    0   -12   0     0   1
5  :  0   -40    0    0   80     0   0   -20   0   0   1
6  : 40     0    0 -360    0     0 220     0   0 -30   0   0 1
7  :  0     0 1120    0    0 -1680   0     0 490   0   0 -42 0   0 1
8  :  0 -2240    0    0 9520     0   0 -5600   0   0 952   0 0 -56 0 0 1
etc.

Crossrefs

Cf. A049404, A066325, A091752.
Row sums are (-1)^n*A252284(n).

Formula

T(n,k) = (-1)^k * (n!) * (Sum_{i=0..floor(k/3)} (-1)^i * binomial((n+k) /3,i) * binomial(n+k-3*i,n)) / (3^((n+k)/3) * ((n+k)/3)!) if (n+k) mod 3 = 0 else 0 with n >= 0 and 0 <= k <= 2*n.
Recurrence: (1) T(n,k) = T(n-1,k-2) - (k+1) * T(n-1,k+1),
(2) T(n,k) = T(n-1,k-2) - 2*(n-1)*T(n-2,k-1) + (n-1)*(n-2)*T(n-3,k),
(3) k*T(n,k) = 2*n*T(n-1,k-2) - n*(n-1)*T(n-2,k-1),
(4) p(n,x) = x^2 * p(n-1,x) - (d/dx) p(n-1,x),
(5) p(n,x) = x^2*p(n-1,x) - 2*(n-1)*x*p(n-2,x) + (n-1)*(n-2)*p(n-3,x),
(6) (d/dx) p(n,x) = 2*n*x*p(n-1,x) - n*(n-1)*p(n-2,x) for n > 0 with initial values T(0,0) = 1 = p(0,x) and T(n,k) = 0 if k < 0 or k > 2*n or (n+k) mod 3 > 0.
T(n,2*n) = 1 for n >= 0.
T(3*n,0) = -T(3*n-1,1) = 2*T(3*n-2,2) = ((3*n)!)/(3^n * (n!)) for n > 0.
The polynomials p(n,x) satisfy for n >= 0 and x complex the differential equation: 0 = (((d/dx)^3) p(n,x)) - 2*x^2*(((d/dx)^2) p(n,x)) + (x^4 + 2*(n-1)*x) * ((d/dx) p(n,x)) - (2*n*x^3-(n+3)*n) * p(n,x).
E.g.f.: Sum_{n>=0} p(n,x)*t^n/(n!) = e^((x^3-(x-t)^3)/3).
((d/dx)^m) p(n,x) = Sum_{i=0..m} (-1)^i * binomial(m,i) * p(m-i,-x) * p(n+i,x) for m,n >= 0 and x complex.
T(3*n-k,k) = A091752(n+1,k+2) for 0 <= k <= 2*n.
(-1)^(n-k) * T(n,3*k-n) = A049404(n,k) for n > 0 and (n+2)/3 <= k <= n.
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