A144646 -id:A144646 - cp's OEIS frontend

A094762 a(n) = Bell(n+1) - 2^n + 1 + n, where Bell(i) is the i-th Bell number A000110(i).

1, 2, 4, 11, 41, 177, 820, 4020, 20900, 115473, 677557, 4211561, 27640354, 190891144, 1382942176, 10480109395, 82864804285, 682076675105, 5832741942932, 51724157711104, 474869815108196, 4506715736350193, 44152005850890065, 445958869286416705

Offset: 0

Jean-Yves Tallet and N. J. A. Sloane, Jan 24 2009

nonn

a(n) is the solution to the following combinatorial problem. Given a set S of n labeled elements, find the number of all subsets of S (2^n) plus the number of partitions of any subset T of S into parts which are not all of size 1 nor of size |T|. This implies that a(n) = 2^n + Sum_{m=3..n} (Bell(m)-2) = Bell(n+1) - 2^n + 1 + n, using the standard recurrence for the Bell numbers (Comtet, Eq. (4a)). - N. J. A. Sloane, Nov 26 2013

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 210.

Cf. A000110, A144646.

with(combinat); [seq(bell(n+1)-2^n+n+1,n=0..30)];

Table[BellB[n+1]-2^n+n+1,{n,0,30}] (* Harvey P. Dale, Apr 24 2018 *)

Also a(n) = 2^n + Sum_{m=3..n} binomial(n,m)*(Bell(m)-2).

cp's OEIS Frontend

A094762 a(n) = Bell(n+1) - 2^n + 1 + n, where Bell(i) is the i-th Bell number A000110(i).

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