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.

A002874 The number of partitions of {1..3n} that are invariant under a permutation consisting of n 3-cycles.

Original entry on oeis.org

1, 2, 8, 42, 268, 1994, 16852, 158778, 1644732, 18532810, 225256740, 2933174842, 40687193548, 598352302474, 9290859275060, 151779798262202, 2600663778494172, 46609915810749130, 871645673599372868, 16971639450858467002, 343382806080459389676
Offset: 0

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Comments

Original name: Sorting numbers.

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

u[n,j] generates for j=1, A000110; j=2, A002872; j=3, this sequence; j=4, A141003; j=5, A036075; j=6, A141004; j=7, A036077. - Wouter Meeussen, Dec 06 2008
Equals column 3 of A162663. - Michel Marcus, Mar 27 2013
Row sums of A294201.

Programs

  • Maple
    S:= series(exp( (exp(3*x) - 4)/3 + exp(x)), x, 31):
    seq(coeff(S,x,j)*j!, j=0..30); # Robert Israel, Oct 30 2015
    # second Maple program:
    a:= proc(n) option remember; `if`(n=0, 1, add((1+
          3^(j-1))*binomial(n-1, j-1)*a(n-j), j=1..n))
        end:
    seq(a(n), n=0..30);  # Alois P. Heinz, Oct 17 2017
  • Mathematica
    u[0,j_]:=1;u[k_,j_]:=u[k,j]=Sum[Binomial[k-1,i-1]Plus@@(u[k-i,j]#^(i-1)&/@Divisors[j]),{i,k}]; Table[u[n,3],{n,0,12}] (* Wouter Meeussen, Dec 06 2008 *)
    mx = 16; p = 3; Range[0, mx]! CoefficientList[ Series[ Exp[ (Exp[p*x] - p - 1)/p + Exp[x]], {x, 0, mx}], x] (* Robert G. Wilson v, Dec 12 2012 *)
    Table[Sum[Binomial[n,k] * 3^k * BellB[k, 1/3] * BellB[n-k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jun 29 2022 *)

Formula

E.g.f.: exp( (exp(3*x) - 4)/3 + exp(x) ).
a(n) ~ exp(exp(3*r)/3 + exp(r) - 4/3 - n) * (n/r)^(n + 1/2) / sqrt((1 + 3*r)*exp(3*r) + (1 + r)*exp(r)), where r = LambertW(3*n)/3 - 1/(1 + 3/LambertW(3*n) + n^(2/3) * (1 + LambertW(3*n)) * (3/LambertW(3*n))^(5/3)). - Vaclav Kotesovec, Jul 03 2022
a(n) ~ (3*n/LambertW(3*n))^n * exp(n/LambertW(3*n) + (3*n/LambertW(3*n))^(1/3) - n - 4/3) / sqrt(1 + LambertW(3*n)). - Vaclav Kotesovec, Jul 10 2022

Extensions

New name from Danny Rorabaugh, Oct 24 2015