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.

A111752 Number of partitions of {1,..,n} into lists with an even number of lists of size 1, where a list means an ordered subset (cf. A000262).

Original entry on oeis.org

1, 0, 3, 6, 49, 300, 2491, 22890, 239457, 2782584, 35595091, 496577070, 7499663953, 121855323876, 2118793593099, 39245026343250, 771255810671041, 16025261292247920, 350956070419872547, 8078570913162379734, 194969375055353840241, 4922311437793379501340
Offset: 0

Views

Author

Vladeta Jovovic, Nov 19 2005; corrected Jun 06 2006

Keywords

Comments

a(n) + A111753(n) = A000262(n). - David Wasserman, Feb 11 2009

Crossrefs

Programs

  • Maple
    b:= proc(n, t) option remember; `if`(n=0, t, add(b(n-j,
         `if`(j=1, 1-t, t))*binomial(n-1, j-1)*j!, j=1..n))
        end:
    a:= n-> b(n, 1):
    seq(a(n), n=0..30);  # Alois P. Heinz, May 10 2016
  • Mathematica
    b[n_, t_] := b[n, t] = If[n == 0, t, Sum[b[n-j, If[j == 1, 1-t, t]] * Binomial[n-1, j-1]*j!, {j, 1, n}]]; a[n_] := b[n, 1]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jan 21 2017, after Alois P. Heinz *)
  • Python
    from sympy.core.cache import cacheit
    from sympy import binomial, factorial as f
    @cacheit
    def b(n, t): return t if n==0 else sum(b(n - j, (1 - t if j==1 else t))*binomial(n - 1, j - 1)*f(j) for j in range(1, n + 1))
    def a(n): return b(n, 1)
    print([a(n) for n in range(51)]) # Indranil Ghosh, Aug 10 2017

Formula

E.g.f.: cosh(x)*exp(x^2/(1-x)). More generally, e.g.f. for number of partitions of {1, 2, ...n} into lists with an even number of lists of size k is cosh(x^k)*exp(x/(1-x)-x^k).
E.g.f.: cosh(x)*exp(x^2/(1-x)) = 1/2*Q(0); Q(k) = 1+((2*x-1)^k)/(1-x/(x+((2*x-1)^k)*(k+1)*(1-x)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 17 2011
a(n) ~ (exp(1)+exp(-1)) * 2^(-3/2) * exp(2*sqrt(n)-n-3/2) * n^(n-1/4) * (1 + (2/(1 + exp(2)) - 5/48)/sqrt(n)). - Vaclav Kotesovec, Jan 21 2017, extended Dec 01 2021

Extensions

More terms from David Wasserman, Feb 11 2009
a(0)=1 prepended by Alois P. Heinz, May 10 2016