cp's OEIS Frontend

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.

A331326 a(n) = n!*[x^n] sinh(x/(1 - x))/(1 - x).

Original entry on oeis.org

0, 1, 4, 19, 112, 801, 6756, 65563, 717760, 8729857, 116570980, 1693096131, 26548383984, 446689827169, 8023582921732, 153192673528651, 3097301219335936, 66095983547942913, 1484384376886189380, 34991710162280602867, 863797053818651591920, 22282392569877969167521
Offset: 0

Views

Author

Peter Luschny, Jan 21 2020

Keywords

Crossrefs

Programs

  • Maple
    gf := sinh(x/(1 - x))/(1 - x): ser := series(gf, x, 22):
    seq(n!*coeff(ser, x, n), n=0..20);
    # Alternative: seq(add(abs(A021009(n, 2*k+1)), k=0..n/2), n=0..21);
    A331326 := proc(n) local S; S := proc(n, k) option remember; `if`(k = 0, 1,
    `if`(k > n, 0, S(n-1, k-1)/k + S(n-1, k))) end: n!*add(S(n, 2*k+1), k=0..n) end:
    seq(A331326(n), n=0..21);
  • Mathematica
    a[n_] := n n! HypergeometricPFQ[{1/2 - n/2, 1 - n/2}, {1, 3/2, 3/2}, 1/4];
    Array[a, 22, 0]
  • PARI
    x='x+O('x^22); concat(0,Vec(serlaplace(sinh(x/(1-x))/(1-x))))
    
  • Python
    def A331326():
        sa, sb, ta, tb, n = 1, 2, 1, 0, 2
        yield 0
        yield ta
        while(True):
            s = 2*n*sb - ((n-1)**2)*sa
            t = 2*(n-1)*tb - ((n-1)**2)*ta
            sa, sb, ta, tb = sb, s, tb, t
            n += 1
            yield (s - t)//2
    a = A331326(); print([next(a) for _ in range(22)])

Formula

a(n) + A331325(n) = A002720(n).
A331325(n) - a(n) = A009940(n).
a(n) = Sum_{k=0..n/2} |A021009(n, 2*k+1)|.
a(n) = Sum_{k=0..n} binomial(n, 2*k+1)*n!/(2*k+1)!.
a(n) = n*n!*hypergeom([1/2 - n/2, 1 - n/2], [1, 3/2, 3/2], 1/4).
(n+1)^2*(n+2)^2*a(n) - 4*(n+2)^3*a(n+1) + (6*n^2+30*n+37)*a(n+2) - 4*(n+3)*a(n+3)+a(n+4) = 0. - Robert Israel, Jan 22 2020
Sum_{n>=0} a(n) * x^n / (n!)^2 = (1/2) * exp(x) * (BesselI(0,2*sqrt(x)) - BesselJ(0,2*sqrt(x))). - Ilya Gutkovskiy, Jul 17 2020
a(n) ~ 2^(-3/2) * exp(2*sqrt(n)-n-1/2) * n^(n+1/4) * (1 + 31/(48*sqrt(n))). - Vaclav Kotesovec, Feb 17 2024