0,3

a(2), a(3), a(5) and a(7) are prime; a(11) is not.

Let S be a set of n elements. First we perform a set partition on S. Let SU be the set of all nonempty subsets of S. As is well known, 2^n - 1 is the number of nonempty subsets of a set with n elements (see A000225). That is, 2^n - 1 is the number of elements |SU| of SU. In the second step, we select k elements from SU. We want to know how many different selections are possible. Let W be the resulting set of selections formed from SU. Then the number of elements |W| of W is |W| = sum((binomial(2^n-1,k)), k=1..2^n-1) = 2^(2^n-1)-1 = A077585. Example: |W(n)| = a(n=2) = 7, because W={[[1, 2]], [[1]], [[1, 2], [1]], [[1, 2], [2], [1]], [[1, 2], [2]], [[2], [1]], [[2]]}. - Thomas Wieder (thomas.wieder(AT)t-online.de), Nov 08 2007

Eric Weisstein's World of Mathematics, Double Mersenne Number

a(n) =A000225(A000225(n)) =A058891(n)-1 =(A001146(n)-2)/2

a(n) = A000225(A000225(n)) = A058891(n)-1 = (A001146(n)-2)/2 = A056220(1+a(n-1)).

a(n) = sum((binomial(2^n-1,k)), k=1..2^n-1) - Thomas Wieder (thomas.wieder(AT)t-online.de), Nov 08 2007

a(5)=2^(2^5-1)-1=2^31-1=2147483647

ZahlDerAuswahlenAusMengeDerZerlegungenEinerMenge:=proc() local n, nend, arg, k, w; nend:=10; for n from 1 to nend do arg:=2^n-1; w[n]:=sum((binomial(arg, k)), k=1..arg); od; print(w[1], w[2], w[3], w[4], w[5], w[6], w[7], w[8], w[9], w[10]); end proc; - Thomas Wieder (thomas.wieder(AT)t-online.de), Nov 08 2007

(PARI) a(n)=if(n<1, 0, -1+2*(1+a(n-1))^2)

(Other) sage: [stirling_number2(2^(n-1), 2) for n in xrange(1, 10)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 27 2009]

Cf. A077586.

Sequence in context: A138523 A034670 A020516 this_sequence A134722 A053713 A077586

Adjacent sequences: A077582 A077583 A077584 this_sequence A077586 A077587 A077588

nonn,new

Henry Bottomley (se16(AT)btinternet.com), Nov 07 2002

Corrected by Lekraj Beedassy (blekraj(AT)yahoo.com), Jan 02 2007