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%I #25 Dec 11 2022 10:43:34
%S 1,5,20,115,1665,82650,12847310,5620114060,6659421195205,
%T 21082748688390045,177217804775828062850,3941798437750184226876305,
%U 231505293200405380457355524620,35848160499603817968830380832049915,14619744406297572472084577939841875791890
%N Expansion of g.f.: exp( Sum_{n>=1} 5*Fibonacci(n^2) * x^n/n ).
%C Moss and Ward prove that this is an integral sequence. - _Peter Bala_, Nov 28 2022
%C Let A(x) be the g.f. for this sequence. Note that the expansion of A(x)^(1/5) = exp( Sum_{n>=1} Fibonacci(n^2) * x^n/n ) does not have integer coefficients.
%H Patrick Moss and Tom Ward, <a href="https://arxiv.org/abs/2011.13068">Fibonacci along even powers is (almost) realizable</a>, arXiv:2011.13068 [math.NT], 2020; Fibonacci Quart. 60 (2022), no. 1, 40-47.
%e G.f.: A(x) = 1 + 5*x + 20*x^2 + 115*x^3 + 1665*x^4 + 82650*x^5 + ...
%e such that
%e log(A(x))/5 = x + 3*x^2/2 + 34*x^3/3 + 987*x^4/4 + 75025*x^5/5 + 14930352*x^6/6 + 7778742049*x^7/7 + ... + Fibonacci(n^2)*x^n/n + ...
%o (PARI) {a(n)=polcoeff(exp(sum(k=1,n,5*fibonacci(k^2)*x^k/k)+x*O(x^n)),n)}
%o for(n=0,16,print1(a(n),", "))
%Y Cf. A054888, A207969, A207970, A207971, A054783, A207834, A211892.
%K nonn,easy
%O 0,2
%A _Paul D. Hanna_, Feb 22 2012
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