%I A107354
%S A107354 1,1,2,7,44,516,11622,512022,44588536,7718806044,2664170119608,
%T A107354 1836214076324153,2529135272371085496,6964321029630556852944,
%U A107354 38346813253279804426846032,422247020982575523983378003936
%N A107354 To compute a(n) we first write down 2^n 1's in a row. Each row takes 
               the right half of the previous row and each element in it equals 
               sum of the elements in the previous row starting at the middle. The 
               single element in the last row is a(n).
%C A107354 Number of subpartitions of partition [1,3,7,...,2^n-1]. - Frank Adams-Watters 
               (FrankTAW(AT)Netscape.net), Mar 11 2006
%C A107354 Can also be computed summing forwards:
%C A107354 1,1
%C A107354 1,2,2,2
%C A107354 1,3,5,7,7,7,7,7
%C A107354 1,4,9,16,23,30,37,44,44,44,44,44,44,44,44,44
%F A107354 a(n) = C(2^(n-1)+n-2, n-1) - Sum_{k=1..n-2} a(k)*C(2^(n-1)-2^k+n-k-1, 
               n-k) for n>2, with a(0)=1, a(1)=1, a(2)=2, where C(n, k)=binomial(n, 
               k). - Paul D. Hanna (pauldhanna(AT)juno.com), May 24 2005
%F A107354 The first number in row 3 is 2^(n-2)+1. - Ralf Stephan (ralf(AT)ark.in-berlin.de), 
               May 24 2005
%F A107354 G.f.: 1/(1-x) = Sum_{n>=0} a(n)*x^n*(1-x)^(2^n-1) (g.f. of subpartitions). 
               - Paul D. Hanna (pauldhanna(AT)juno.com), Jul 03 2006
%F A107354 G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1+x)^(2^n+n). - Paul D. Hanna (pauldhanna(AT)juno.com), 
               Jul 03 2006
%e A107354 For example, for n=4 the array looks like this:
%e A107354 1..1..1..1..1..1..1..1..1..1..1..1..1..1..1..1
%e A107354 ........................1..2..3..4..5..6..7..8
%e A107354 ....................................5.11.18.26
%e A107354 .........................................18.44
%e A107354 ............................................44
%e A107354 Therefore a(n)=44.
%e A107354 At n=5, we can illustrate the recurrence by:
%e A107354 a(5) = 516 = C(19, 4) - ( 1*C(17, 4) + 2*C(14, 3) + 7*C(9, 2)
%e A107354 )
%e A107354 = C(16+4-1, 4) - (1*C(16-2+4-1, 4) + 2*C(16-4+3-1, 3) + 7*C(16-8+2-1, 
               2) ).
%t A107354 f[n_] := If[n == 0, 1, Binomial[2^(n - 1) + n - 2, n - 1] - Sum[ f[k]*Binomial[2^(n 
               - 1) - 2^k + n - k - 1, n - k], {k, n - 2}]]; Table[ f[n], {n, 0, 
               15}] (from Robert G. Wilson v (rgwv(AT)rgwv.com), May 25 2005)
%o A107354 (PARI) {a(n)=if(n==0,1,binomial(2^(n-1)+n-2,n-1)- sum(k=1,n-2,a(k)*binomial(2^(n-1)-2^k+n-k-1,
               n-k)))} (Hanna)
%o A107354 (PARI) {a(n)=polcoeff(x^n-sum(k=0, n-1, a(k)*x^k*(1-x+x*O(x^n))^(2^k-1) 
               ), n)} (Hanna)
%Y A107354 Cf. A105996; variants: A109055 - A109061; subpartitions defined: A115728, 
               A115729.
%Y A107354 Sequence in context: A001046 A158257 A153522 this_sequence A006118 A083670 
               A108240
%Y A107354 Adjacent sequences: A107351 A107352 A107353 this_sequence A107355 A107356 
               A107357
%K A107354 nonn,nice
%O A107354 0,3
%A A107354 Max Alekseyev (maxale(AT)gmail.com), May 24 2005
%E A107354 Edited by Paul Hanna, Jul 03 2006