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G.f. A(x) satisfies: A(x) = 1 - x * A(x/(1 - x)) / (1 - x)^3.
3

%I #12 Dec 01 2022 08:57:59

%S 1,-1,-2,0,10,25,-11,-301,-1040,-60,17770,95359,146701,-1513837,

%T -14210258,-53101500,91834402,2739189073,19172894377,46384729811,

%U -498471972128,-7229201676480,-45007184571062,-40076612769641,2435999270437801,30321258115161275,180120147363157438

%N G.f. A(x) satisfies: A(x) = 1 - x * A(x/(1 - x)) / (1 - x)^3.

%H Seiichi Manyama, <a href="/A346053/b346053.txt">Table of n, a(n) for n = 0..592</a>

%F a(n+1) = -Sum_{k=0..n} binomial(n+2,k+2) * a(k).

%t nmax = 26; A[_] = 0; Do[A[x_] = 1 - x A[x/(1 - x)]/(1 - x)^3 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%t a[0] = 1; a[n_] := a[n] = -Sum[Binomial[n + 1, k + 2] a[k], {k, 0, n - 1}]; Table[a[n], {n, 0, 26}]

%o (SageMath)

%o @CachedFunction

%o def a(n): # a = A346053

%o if (n==0): return 1

%o else: return (-1)*sum(binomial(n+1, k+2)*a(k) for k in range(n))

%o [a(n) for n in range(51)] # _G. C. Greubel_, Dec 01 2022

%Y Cf. A000587, A014619, A045501.

%K sign

%O 0,3

%A _Ilya Gutkovskiy_, Jul 02 2021