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.

A331325 a(n) = n!*[x^n] cosh(x/(1-x))/(1-x).

Original entry on oeis.org

1, 1, 3, 15, 97, 745, 6571, 65359, 723969, 8842257, 118091251, 1712261551, 26786070433, 449634481465, 8059974923547, 153634497337455, 3102367733191681, 66145005096272929, 1484586887025099619, 34983117545622446287, 863397428225495045601, 22269844592814969946761
Offset: 0

Views

Author

Peter Luschny, Jan 21 2020

Keywords

Crossrefs

Programs

  • Maple
    gf := cosh(x/(1 - x))/(1 - x): ser := series(gf, x, 22):
    seq(n!*coeff(ser, x, n), n=0..21);
    # Alternative: seq(add(abs(A021009(n, 2*k)), k=0..n/2), n=0..21);
    A331325 := 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), k=0..n) end:
    seq(A331325(n), n=0..21);
  • Mathematica
    a[n_] := n! HypergeometricPFQ[{1/2 - n/2, -n/2}, {1, 1/2, 1/2}, 1/4];
    Array[a, 22, 0]
  • PARI
    x='x+O('x^22); Vec(serlaplace(cosh(x/(1-x))/(1-x)))
    
  • Python
    def A331325():
        sa, sb, ta, tb, n = 1, 2, 1, 0, 2
        yield sa
        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 = A331325(); print([next(a) for _ in range(22)])

Formula

a(n) + A331326(n) = A002720(n).
a(n) - A331326(n) = A009940(n).
a(n) = Sum_{k=0..n/2} |A021009(n, 2*k)|.
a(n) = Sum_{k=0..n} binomial(n, 2*k)*n!/(2*k)!.
a(n) = n!*hypergeom([1/2 - n/2, -n/2], [1/2, 1/2, 1], 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 23 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 18 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