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!).
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
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
- Pierre Humbert, Monographie des polynômes de Kummer, Nouvelles annales de mathématiques, journal des candidats aux écoles polytechnique et normale, Serie 5, Volume 1 (1922), pp. 81-92.
- E. E. Kummer, Über die hypergeometrische Reihe, Journal für die reine und angewandte Mathematik 15 (1836): 39-83.
- T. Mansour, M. Schork and M. Shattuck, The Generalized Stirling and Bell Numbers Revisited, Journal of Integer Sequences, Vol. 15 (2012), #12.8.3.
- Ladislav Truksa, Hypergeometric orthogonal systems of polynomials III, Aktuárské vědy, Vol. 2 (1931), No. 4, 177-203, (see p.200).
Crossrefs
Programs
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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);
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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)).
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.
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