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%I #8 Sep 20 2013 11:34:42
%S 1,2,3,4,8,8,20,16,48,32,112,64,256,128,576,256,1280,512,2816,1024,
%T 6144,2048,13312,4096,28672,8192,61440,16384,131072,32768,278528,
%U 65536,589824,131072,1245184,262144,2621440,524288,5505024
%N Row sums of A211226.
%C The odd-indexed terms of the sequence a(2*n-1) count the compositions of n+1, while the even-indexed terms a(2*n) count the total number of parts in the composition of n+1. Compare with A211228.
%F a(n) = sum {k = 0..n } f(n)/(f(k)*f(n-k)), where f(n) := (floor(n/2))!.
%F a(2*n-1) = 2^n = A000079(n); a(2*n) = (n+2)*2^(n-1) = A001792(n).
%F O.g.f.: (1+2*x-x^2-4*x^3)/(1-2*x^2)^2 = 1 + 2*x + 3*x^2 + 4*x^3 + 8*x^4 + ....
%F E.g.f.: cosh(sqrt(2)*x) + (4+x)/(2*sqrt(2))*sinh(sqrt(2)*x) = 1 + 2*x + 3*x^2/2! + 4*x^3/3! + 8*x^4/4! + .....
%e The four compositions of 3 are 1+1+1, 1+2, 2+1 and 3 having 8 parts in total. Hence a(3) = 4 and a(4) = 8.
%Y Cf. A000079, A001792, A211226.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Apr 05 2012