A353050 G.f. A(x) satisfies: 0 = Sum_{n>=1} (Lucas(n) - A(x))^n * x^n/n, where Lucas(n) = A000204(n).

1, 2, 5, 298, 18949, 3962150, 1916564344, 2501114025582, 8336852053702202, 73027618049882652700, 1666798946804859125492899, 99738726494828465657124210156, 15634873312495144092899303952245929, 6430416165010536428917103922818349814504

Offset: 0

Author

Paul D. Hanna, Apr 26 2022

Keywords

• nonn

Comments

It is conjectured that this is an integer sequence.

This sequence is a special case of a more general conjecture: 0 = Sum_{n>=1} (L(n) - F(x))^n * x^n/n is satisfied by an integer series F(x) if exp( Sum_{n>=1} L(n) * x^n/n ) yields an integer series.

Examples

			G.f.: A(x) = 1 + 2*x + 5*x^2 + 298*x^3 + 18949*x^4 + 3962150*x^5 + 1916564344*x^6 + 2501114025582*x^7 + 8336852053702202*x^8 + ...
where
0 = (1 - A(x))*x + (3 - A(x))^2*x^2/2 + (4 - A(x))^3*x^3/3 + (7 - A(x))^4*x^4/4 + (11 - A(x))^5*x^5/5 + (18 - A(x))^6*x^6/6 + (29 - A(x))^7*x^7/7 + (47 - A(x))^8*x^8/8 + (76 - A(x))^9*x^9/9 + ... + (Lucas(n) - A(x))^n*x^n/n + ...
Related series.
exp( Sum_{n>=1} A(x)^n * x^n/n ) = 1/(1 - x*A(x)) = 1 + x + 3*x^2 + 10*x^3 + 319*x^4 + 19601*x^5 + 4002282*x^6 + 1924629400*x^7 + 2504975492897*x^8 + 8341867813691252*x^9 + ...
exp( Sum_{n>=1} Lucas(n)^n * x^n/n ) = 1 + x + 5*x^2 + 26*x^3 + 634*x^4 + 32928*x^5 + 5704263*x^6 + 2470113915*x^7 + 2978904483553*x^8 + 9401949327631932*x^9 + ... + A156216(n)*x^n + ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..100

Crossrefs

Cf. A156216, A316363, A000204.

Programs

PARI

{Lucas(n) = fibonacci(n-1) + fibonacci(n+1)}
{a(n) = my(A=[1]); for(i=1,n, A=concat(A,0);
A[#A] = polcoeff( sum(m=1,#A, (Lucas(m) - Ser(A))^m*x^m/m), #A));A[n+1]}
for(n=0,20,print1(a(n),", "))