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%I #15 Jul 29 2019 13:16:18
%S 1,1,3,19,251,6843,381851,43357211,9976746651,4639483488923,
%T 4351708606681243,8221479626141796507,31252321079882850259099,
%U 238835886863534101328335003,3667031594654877566958673359003,113055325655546855868908521812586651
%N Coefficients associated with power series solution of g'(x) = g(x/2) e^x at x=0.
%C The series g(x) := Sum_{n>=0} a(n) * 2^((-n*n+n)/2) * x^n / n! satisfies g'(x) = g(x/2) e^x.
%C The denominators of the power series can be found at A006125. - _Lewis Chen_, Apr 28 2019
%H FiveThirtyEight, <a href="https://fivethirtyeight.com/features/how-many-earthlings-would-survive-63-thanos-snaps/">The Riddler (Apr. 26, 2019)</a> (Riddler Express), with <a href="https://fivethirtyeight.com/features/are-you-the-best-warlord/">solution</a>
%F a(n+1) = Sum_{n=0..k} a(k)*binomial(n, k)*2^( (n*n+n - k*k-k)/2 ).
%o (PARI) {a(n) = if( n<1, n==0, n--; sum(k=0, n, a(k) * binomial(n, k) * 2^( (n*n+n - k*k-k)/2 )))};
%Y Cf. A006125.
%K nonn,frac
%O 0,3
%A _Michael Somos_, Jun 04 2018
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