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%I #71 Oct 19 2023 12:52:45
%S 1,2,12,104,1200,17312,299712,6053504,139733760,3628677632,
%T 104701504512,3323151509504,115063060869120,4316023589937152,
%U 174347763227738112,7545919601962287104,348366745238330081280,17087957176042900815872,887497598764802460352512
%N Number of set partitions of {1,2,...,n} with labeled blocks and a (possibly empty) subset of designated elements in each block.
%H Vincenzo Librandi, <a href="/A216794/b216794.txt">Table of n, a(n) for n = 0..200</a>
%H José A. Adell, Beáta Bényi, Venkat Murali, and Sithembele Nkonkobe, <a href="https://doi.org/10.22108/toc.2022.130037.1894">Generalized Barred Preferential Arrangements</a>, Transactions on Combinatorics (2022).
%H Sithembele Nkonkobe, Venkat Murali, and Béata Bényi, <a href="https://arxiv.org/abs/1907.08944">Generalised Barred Preferential Arrangements</a>, arXiv:1907.08944 [math.CO], 2019.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Polylogarithm.html">Polylogarithm</a>.
%F E.g.f.: 1/(2 - exp(2*x)).
%F 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
%F 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
%F 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
%F a(n) ~ n! * (2/log(2))^n/log(4). - _Vaclav Kotesovec_, Sep 24 2013
%F 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
%F From _Vladimir Reshetnikov_, Oct 31 2015: (Start)
%F a(n) = (-1)^(n+1)*(Li_{-n}(sqrt(2)) + Li_{-n}(-sqrt(2)))/4, where Li_n(x) is the polylogarithm.
%F Li_{-n}(sqrt(2)) = (-1)^(n+1)*(2*a(n) + A080253(n)*sqrt(2)).
%F (End)
%F a(n) = 2^(n-1)*(Li_{-n}(1/2) + 0^n) with 0^0=1. - _Peter Luschny_, Nov 03 2015
%F From _Peter Bala_, Oct 18 2023: (Start)
%F a(n) = 2^n * A000670(n)
%F Inverse binomial transform of A080253.
%F The sequence is the first column of the array (2*I - P^2)^(-1), where P denotes Pascal's triangle A007318. (End)
%p a := n -> 2^(n-1)*(polylog(-n, 1/2)+`if`(n=0,1,0)):
%p seq(round(evalf(a(n),32)), n=0..18); # _Peter Luschny_, Nov 03 2015
%p # second Maple program:
%p a:= proc(n) option remember; `if`(n=0, 1, add(
%p a(n-j)*binomial(n, j)*2^j, j=1..n))
%p end:
%p seq(a(n), n=0..20); # _Alois P. Heinz_, Oct 04 2019
%t nn=25;a=Exp[2x]-1;Range[0,nn]!CoefficientList[Series[1/(1-a),{x,0,nn}],x]
%t Round@Table[(-1)^(n+1) (PolyLog[-n, Sqrt[2]] + PolyLog[-n, -Sqrt[2]])/4, {n, 0, 20}] (* _Vladimir Reshetnikov_, Oct 31 2015 *)
%o (Sage)
%o def A216794(n):
%o 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))
%o [A216794(n) for n in range(18)] # _Peter Luschny_, Jul 22 2014
%o (PARI) a(n) = 2^(n-1)*(polylog(-n, 1/2) + 0^n); \\ _Michel Marcus_, May 30 2018
%Y Cf. A000556, A000670, A006153, A055882, A080253.
%K nonn,easy
%O 0,2
%A _Geoffrey Critzer_, Sep 16 2012
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