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G.f. satisfies A(x) = 1 + x*A(x) / (1 + x)^5.
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%I #14 Aug 21 2023 08:23:02

%S 1,1,-4,6,6,-49,95,24,-592,1417,-414,-6809,20142,-14831,-73353,274761,

%T -311105,-715647,3607624,-5463428,-5785294,45588556,-87189477,

%U -25565196,552659892,-1305250324,340413165,6379267117,-18606431142,13202513476,69064770845

%N G.f. satisfies A(x) = 1 + x*A(x) / (1 + x)^5.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (-4,-10,-10,-5,-1).

%F G.f.: A(x) = 1/( 1 - x/(1+x)^5 ).

%F a(n) = -4*a(n-1) - 10*a(n-2) - 10*a(n-3) - 5*a(n-4) - a(n-5) for n > 5.

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

%t LinearRecurrence[{-4, -10, -10, -5, -1}, {1, 1, -4, 6, 6, -49}, 1 + 30] (* _Robert P. P. McKone_, Aug 21 2023 *)

%o (PARI) a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n+4*k-1, n-k));

%Y Cf. A106510, A127896, A365083.

%Y Cf. A079675.

%K sign,easy

%O 0,3

%A _Seiichi Manyama_, Aug 21 2023