%I
%S 1,2,5,8,16,24,40,56,88,120,176,232,328,424,576,728,968,1208,1568,
%T 1928,2464,3000,3768,4536,5632,6728,8248,9768,11864,13960,16784,19608,
%U 23400,27192,32192,37192,43760,50328,58824,67320,78280,89240,103200,117160
%N Given M = triangle A122196 as an infinite lower triangular matrix, this sequence is Lim_{n=1..inf.} M^n.
%C A171238 also = polcoeff: (1,2,3,...)*(1,0,2,0,5,0,8,0,16,...).
%C Number of binary partitions of n into two kinds of parts.  _Joerg Arndt_, Feb 26 2015
%C Let the nth convolution power of the sequence = B, with C = the aerated variant of B. It appears that B/C = the binomial sequence starting (1, 2n,...). Example: The sequence squared = (1, 4, 14, 36, 89, 192,...) = B; with C = (1, 0, 4, 0, 14, 0, 36,...). Then B/C = A000292: (1, 4, 10, 20, 35, 56,...).  _Gary W. Adamson_, Aug 15 2016
%H Vincenzo Librandi, <a href="/A171238/b171238.txt">Table of n, a(n) for n = 1..1000</a>
%F Given M = triangle A122196 as an infinite lower triangular matrix, this sequence is Lim_{n=1..inf.}, a leftshifted vector considered as a sequence.
%F From _Wolfdieter Lang_, Jul 15 2010: (Start)
%F O.g.f.: x*Q(x) with Q(x)*(1x)^2 = Q(x^2), for the eigensequence M*Q = Q with the column o.g.f.s (x^(2*m))/(1x)^2, m>=0, of M.
%F Recurrence for b(n):=a(n+1): b(n)=0 if n<0, b(0)=1; if n even then b(n) = b(n/2)+2*b(n1)b(n2), else b(n) = 2*b(n1) b(n2). (End)
%F G.f.: 1/((1x)*(1x^2)*(1x^4)* ... *(1 x^(2^k))* ...)^2.  _Robert G. Wilson v_, May 11 2012
%F Convolution square of A018819.  _Michael Somos_, Mar 28 2014
%e G.f. = x + 2*x^2 + 5*x^3 + 8*x^4 + 16*x^5 + 24*x^6 + 40*x^7 + 56*x^8 + ...
%t CoefficientList[ Series[ 1/ Product[1  x^(2^i), {i, 0, 6}]^2, {x, 0, 60}], x] (* _Robert G. Wilson v_, May 11 2012 *)
%Y Cf. A018819.
%Y Cf. A000292
%K nonn,easy
%O 1,2
%A _Gary W. Adamson_, Dec 05 2009
%E More terms from _Wolfdieter Lang_, Jul 15 2010
