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A144691 Limit of the coefficient of x^(2^m+n) in B(x)^(n+1)/(n+1) as m grows, where B(x) = Sum_{k>=0} x^(2^k). 5

%I #22 Mar 30 2012 18:37:14

%S 1,1,2,4,26,106,816,4292,90162,715138,10275886,87498566,1944309280,

%T 20988667064,380829128200,4301687654136,219999839271970,

%U 3375111608092354,90438559754079802,1341646116200287978,52342848299405537114,921821277222438350170

%N Limit of the coefficient of x^(2^m+n) in B(x)^(n+1)/(n+1) as m grows, where B(x) = Sum_{k>=0} x^(2^k).

%H Max Alekseyev, <a href="/A144691/b144691.txt">Table of n, a(n) for n = 0..27</a>

%F a(n) = A144690(n)/(n+1).

%F G.f. A(x) satisfies: A(x/(1+x))/(1+x) is an even function; i.e., the inverse binomial transform yields A202582.

%e G.f.: A(x) = 1 + x + 2*x^2 + 4*x^3 + 26*x^4 + 106*x^5 + 816*x^6 +...

%e A(x/G(x)) = G(x) = x/Series_Reversion[x*A(x)], where

%e G(x) = 1 + x + x^2 + 17*x^4 + 408*x^6 + 69473*x^8 + 6018928*x^10 +...

%e and G(x) appears to continue with only even powers of x (cf. A144692).

%e The inverse binomial transform forms the g.f. of A202582:

%e A(x/(1+x))/(1+x) = 1 + x^2 + 19*x^4 + 515*x^6 + 74383*x^8 + 6816465*x^10 +...+ A202582(n)*x^n +...

%o (PARI) { a(n) = local(m=n+log(n+.5)\log(2), B=sum(k=0,m,x^(2^k))); if(n<0, 0, polcoeff((B+O(x^(2^m+n+1)))^(n+1)/(n+1),2^m+n)) }

%Y Cf. A007178, A144690, A144692, A202582.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Oct 10 2008

%E a(14), a(15) corrected and a(16)-a(23) added by _Max Alekseyev_, May 03 2011

%E a(24)-a(27) in b-file from _Max Alekseyev_, Dec 19 2011

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