A349048 - OEIS

%I #10 Nov 07 2021 11:18:10

%S 1,1,1,1,0,-2,-5,-9,-12,-10,3,35,91,163,215,163,-136,-858,-2107,-3675,

%T -4639,-2879,5161,23741,54910,91988,108843,47483,-186582,-700420,

%U -1527461,-2440985,-2656442,-507076,6617735,21456279,44213835,67037683,65541879,-9699085,-232548686

%N G.f. A(x) satisfies: A(x) = 1 / (1 - x + x^4 * A(x)).

%H Seiichi Manyama, <a href="/A349048/b349048.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: (-1 + x + sqrt((1 - x)^2 + 4*x^4)) / (2*x^4).

%F a(0) = 1; a(n) = a(n-1) - Sum_{k=0..n-4} a(k) * a(n-k-4).

%F a(n) = Sum_{k=0..floor(n/4)} (-1)^k * binomial(n-2*k,2*k) * Catalan(k).

%F a(n) = F([(1-n)/4, (2-n)/4, (3-n)/4, -n/4], [2, (1-n)/2, -n/2], -64), where F is the generalized hypergeometric function. - _Stefano Spezia_, Nov 07 2021

%t nmax = 40; A[_] = 0; Do[A[x_] = 1/(1 - x + x^4 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%t a[0] = 1; a[n_] := a[n] = a[n - 1] - Sum[a[k] a[n - k - 4], {k, 0, n - 4}]; Table[a[n], {n, 0, 40}]

%t Table[Sum[(-1)^k Binomial[n - 2 k, 2 k] CatalanNumber[k], {k, 0, Floor[n/4]}], {n, 0, 40}]

%Y Cf. A000108, A007440, A023426, A100223, A214649, A343773, A349047.

%K sign

%O 0,6

%A _Ilya Gutkovskiy_, Nov 06 2021