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%I #11 Feb 02 2020 17:39:29
%S 1,1,3,43,1625,192785,73792371,94005141667,408909577044065,
%T 6204433373664395569,334203804752658372354515,
%U 64828498485572980097719939179,45811084061472137471487315433296153,119028111984311982345314987179033877373025,1145664208319965667452046935744516601565935434531
%N Inverse binomial transform of A014070: a(n) = Sum_{k=0..n} (-1)^(n-k)*C(n,k)*C(2^k,k).
%H Andrew Howroyd, <a href="/A136648/b136648.txt">Table of n, a(n) for n = 0..50</a>
%F G.f.: A(x) = (1/(1+x))*Sum_{n>=0} [log(1 + (2^n+1)*x) - log(1+x)]^n / n!.
%F a(n) ~ 2^(n^2) / n!. - _Vaclav Kotesovec_, Jul 02 2016
%t Table[Sum[(-1)^(n-k)*Binomial[n,k]*Binomial[2^k,k], {k, 0, n}], {n, 0, 15}] (* _Vaclav Kotesovec_, Jul 02 2016 *)
%o (PARI) {a(n)=sum(k=0,n,(-1)^(n-k)*binomial(n,k)*binomial(2^k,k))}
%o (PARI) /* Using the g.f.: */ {a(n)=local(X=x+x*O(x^n));polcoeff(sum(k=0,n,(log(1+2^k+1)*X)-log(1+X))^k/k!)/(1+X),n)}
%Y Cf. A014070 (C(2^n, n)), A134174.
%K nonn
%O 0,3
%A _Paul D. Hanna_ and _Vladeta Jovovic_, Jan 21 2008
%E Edited by _Charles R Greathouse IV_, Oct 28 2009
%E Terms a(13) and beyond from _Andrew Howroyd_, Feb 02 2020
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