A216794 Number of set partitions of {1,2,...,n} with labeled blocks and a (possibly empty) subset of designated elements in each block.

1, 2, 12, 104, 1200, 17312, 299712, 6053504, 139733760, 3628677632, 104701504512, 3323151509504, 115063060869120, 4316023589937152, 174347763227738112, 7545919601962287104, 348366745238330081280, 17087957176042900815872, 887497598764802460352512

Offset: 0

Author

Geoffrey Critzer, Sep 16 2012

Keywords

• nonn
• easy

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• José A. Adell, Beáta Bényi, Venkat Murali, and Sithembele Nkonkobe, Generalized Barred Preferential Arrangements, Transactions on Combinatorics (2022).
• Sithembele Nkonkobe, Venkat Murali, and Béata Bényi, Generalised Barred Preferential Arrangements, arXiv:1907.08944 [math.CO], 2019.
• Eric Weisstein's World of Mathematics, Polylogarithm.

Crossrefs

Cf. A000556, A000670, A006153, A055882, A080253.

Programs

Maple

a := n -> 2^(n-1)*(polylog(-n, 1/2)+`if`(n=0,1,0)):
seq(round(evalf(a(n),32)), n=0..18); # Peter Luschny, Nov 03 2015
# second Maple program:
a:= proc(n) option remember; `if`(n=0, 1, add(
      a(n-j)*binomial(n, j)*2^j, j=1..n))
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, Oct 04 2019

Mathematica

nn=25;a=Exp[2x]-1;Range[0,nn]!CoefficientList[Series[1/(1-a),{x,0,nn}],x]
Round@Table[(-1)^(n+1) (PolyLog[-n, Sqrt[2]] + PolyLog[-n, -Sqrt[2]])/4, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 31 2015 *)

PARI

a(n) = 2^(n-1)*(polylog(-n, 1/2) + 0^n); \\ Michel Marcus, May 30 2018

Sage

def A216794(n):
    return 2^n*add(add((-1)^(j-i)*binomial(j,i)*i^n for i in range(n+1)) for j in range(n+1))
[A216794(n) for n in range(18)] # Peter Luschny, Jul 22 2014

Formula

E.g.f.: 1/(2 - exp(2*x)).

E.g.f.: 1 + 2*x/(G(0) - 2*x) where G(k) = 2*k+1 - x*2*(2*k+1)/(2*x + (2*k+2)/(1 + 2*x/G(k+1))); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 26 2012

E.g.f.: 1 + 2*x/( G(0) - 2*x ) where G(k) = 1 - 2*x/(1 + (1*k+1)/G(k+1)); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 02 2013

G.f.: 1/G(0) where G(k) = 1 - x*(2*k+2)/( 1 - 4*x*(k+1)/G(k+1) ); (continued fraction ). - Sergei N. Gladkovskii, Mar 23 2013

a(n) ~ n! * (2/log(2))^n/log(4). - Vaclav Kotesovec, Sep 24 2013

G.f.: T(0)/(1-2*x), where T(k) = 1 - 8*x^2*(k+1)^2/( 8*x^2*(k+1)^2 - (1-2*x-6*x*k)*(1-8*x-6*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 14 2013

From Vladimir Reshetnikov, Oct 31 2015: (Start)

a(n) = (-1)^(n+1)*(Li_{-n}(sqrt(2)) + Li_{-n}(-sqrt(2)))/4, where Li_n(x) is the polylogarithm.

Li_{-n}(sqrt(2)) = (-1)^(n+1)*(2*a(n) + A080253(n)*sqrt(2)).

(End)

a(n) = 2^(n-1)*(Li_{-n}(1/2) + 0^n) with 0^0=1. - Peter Luschny, Nov 03 2015

From Peter Bala, Oct 18 2023: (Start)

a(n) = 2^n * A000670(n)

Inverse binomial transform of A080253.

The sequence is the first column of the array (2*I - P^2)^(-1), where P denotes Pascal's triangle A007318. (End)