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A207970 G.f.: exp( Sum_{n>=1} 5*Fibonacci(n)^6 * x^n/n ).

%I A207970 #23 Oct 18 2020 05:21:30
%S A207970 1,5,15,140,1505,21875,319620,4936985,77358485,1236083870,19982821875,
%T A207970 326511608255,5379199407890,89249496596015,1489580814490755,
%U A207970 24988546214618750,421055477328447620,7122346563647277860,120891417096833214485,2058225554792946621495
%N A207970 G.f.: exp( Sum_{n>=1} 5*Fibonacci(n)^6 * x^n/n ).
%C A207970 Conjecture: exp( Sum_{n>=1} 5*Fibonacci(n)^(2*k) * x^n/n ) is an integer series for integers k >= 0.
%C A207970 Note that exp( Sum_{n>=1} 5*Fibonacci(n)^(2*k+1) * x^n/n ) is not an integer series for integers k.
%C A207970 Note that exp( Sum_{n>=1} Fibonacci(n)^(2*k) * x^n/n ) is not an integer series for integers k.
%F A207970 The o.g.f. A(x) = 1 + 5*x + 15*x^2 + 140*x^3 + ... is an algebraic function: A(x)^25 = ( (1 + 2*x + x^2)^10*(1 + 7*x + x^2)^6 )/( (1 - 3*x + x^2)^15*(1 - 18*x + x^2) ). Cf. A203806. - _Peter Bala_, Apr 03 2014
%F A207970 a(n) ~ 2^(17/25) * 5^(13/50) * phi^(6*n) / (Gamma(1/25) * 3^(3/5) * n^(24/25)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - _Vaclav Kotesovec_, Oct 18 2020
%e A207970 G.f.: A(x) = 1 + 5*x + 15*x^2 + 140*x^3 + 1505*x^4 + 21875*x^5 + 319620*x^6 + ...
%e A207970 such that
%e A207970 log(A(x))/5 = x + x^2/2 + 2^6*x^3/3 + 3^6*x^4/4 + 5^6*x^5/5 + 8^6*x^6/6 + 13^6*x^7/7 + ... + Fibonacci(n)^6*x^n/n + ...
%o A207970 (PARI) {a(n)=polcoeff(exp(sum(k=1,n,5*fibonacci(k)^6*x^k/k)+x*O(x^n)),n)}
%o A207970 for(n=0,31,print1(a(n),", "))
%Y A207970 Cf. A054888, A207969, A207971, A207972, A207834, A207834. A077916, A203806.
%K A207970 nonn
%O A207970 0,2
%A A207970 _Paul D. Hanna_, Feb 22 2012