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G.f.: exp( Sum_{n>=1} 4^A001511(n) * x^n/n ), where 2^A001511(n) is the highest power of 2 that divides 2*n.
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%I #29 Oct 30 2012 01:07:57

%S 1,4,16,44,128,308,752,1628,3584,7268,14864,28556,55296,102036,189168,

%T 337084,603136,1044676,1814288,3064556,5188352,8578548,14205936,

%U 23041308,37420800,59680548,95265552,149620812,235161216,364301652,564627952,863725948,1321756672

%N G.f.: exp( Sum_{n>=1} 4^A001511(n) * x^n/n ), where 2^A001511(n) is the highest power of 2 that divides 2*n.

%C Compare g.f. to the g.f. of binary partitions (A000123):

%C exp( Sum_{n>=1} 2^A001511(n) * x^n/n ).

%F Self-convolution of A162581.

%e G.f.: A(x) = 1 + 4*x + 16*x^2 + 44*x^3 + 128*x^4 + 308*x^5 + 752*x^6 +...

%e where

%e log(A(x)) = 4^1*x + 4^2*x^2/2 + 4^1*x^3/3 + 4^4*x^4/4 + 4^1*x^5/5 + 4^2*x^6/6 + 4^1*x^7/7 + 4^4*x^8/8 + 4^1*x^9/9 + 4^2*x^10/10 + 4^1*x^11/11 + 4^4*x^12/12 +...+ 4^A001511(n)*x^n/n +...

%o (PARI) {a(n)=polcoeff(exp(sum(m=1,n,4^valuation(2*m,2)*x^m/m)+x*O(x^n)),n)}

%o for(n=0,31,print1(a(n),", "))

%Y Cf. A162581, A180591, A001511.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Oct 30 2012