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%I #16 Apr 05 2020 15:53:02
%S 1,2,8,20,114,288,1156,3256,23464,59716,243212,699216,3659988,
%T 10265800,42353168,128163440,1127515970,2858004752,11768578868,
%U 34294832344,180335471424,513911386232,2137413847256,6572758142016,41948816796852
%N G.f.: A(x) = exp( Sum_{n>=1} C(2n,n)*A006519(n) * x^n/n ), where A006519(n) = highest power of 2 dividing n.
%C Compare g.f. to the g.f. of the Catalan numbers: exp( Sum_{n>=1} C(2n,n)*x^n/n ), where C(2n,n) form the central binomial coefficients (A000984).
%H G. C. Greubel, <a href="/A162585/b162585.txt">Table of n, a(n) for n = 0..1000</a>
%e G.f.: A(x) = 1 + 2*x + 6*x^2 + 10*x^3 + 146*x^4 + 282*x^5 + 826*x^6 + ...
%e log(A(x)) = 2*x + 12*x^2/2 + 20*x^3/3 + 280*x^4/4 + 252*x^5/5 + 1848*x^6/6 + ... + C(2n,n)*A006519(n)*x^n/n + ...
%t nmax=50; CoefficientList[Series[Exp[Sum[2^(IntegerExponent[k, 2])*Binomial[2*k, k]*q^k/k, {k,nmax+3}]], {q,0,nmax}], q] (* _G. C. Greubel_, Jul 04 2018 *)
%o (PARI) {a(n)=local(L=sum(m=1,n,2^valuation(m,2)*binomial(2*m,m)*x^m/m)+x*O(x^n));polcoeff(exp(L),n)}
%Y Cf. A000108, A000123, A000984, A006519.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jul 06 2009
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