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E.g.f.: exp(x/(1-5*x)).
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%I #30 Oct 19 2024 17:39:07

%S 1,1,11,181,3961,108101,3532651,134415961,5834249681,284391878761,

%T 15378011541451,913297438474301,59086483931657161,4135583008765323181,

%U 311324086814794408811,25079793551003791168801,2152597370423901820231201,196089415332225446044417361

%N E.g.f.: exp(x/(1-5*x)).

%C For k = 2,3,4,... the difference a(n+k) - a(n) is divisible by k.

%H Ludovic Schwob, <a href="/A321848/b321848.txt">Table of n, a(n) for n = 0..199</a>

%H Norihiro Nakashima, Shuhei Tsujie, <a href="https://arxiv.org/abs/1904.09748">Enumeration of Flats of the Extended Catalan and Shi Arrangements with Species</a>, arXiv:1904.09748 [math.CO], 2019.

%F a(n) = Sum_{k=0..n} 5^(n-k)*(n!/k!)*binomial(n-1, k-1).

%F Recurrence: a(n) = (10*n-9)*a(n-1) - 25*(n-2)*(n-1)*a(n-2).

%F a(n) ~ n! * exp(2*sqrt(n/5) - 1/10) * 5^(n - 1/4) / (2 * sqrt(Pi) * n^(3/4)). - _Vaclav Kotesovec_, Nov 21 2018

%p seq(coeff(series(factorial(n)*exp(x/(1-5*x)),x,n+1), x, n), n = 0 .. 17); # _Muniru A Asiru_, Nov 24 2018

%t a[n_] := Sum[5^(n - k)*n!/k!*Binomial[n - 1, k - 1], {k, 0, n}]; Array[a, 20, 0] (* or *) a[0] = a[1] = 1; a[n_] := a[n] = (10 n - 9)*a[n - 1] - 25(n - 2)(n - 1)*a[n - 2]; Array[a, 20, 0] (* _Amiram Eldar_, Nov 19 2018 *)

%t With[{nn=20},CoefficientList[Series[Exp[x/(1-5x)],{x,0,nn}],x] Range[0,nn]!] (* _Harvey P. Dale_, Oct 19 2024 *)

%o (PARI) my(x='x + O('x^20)); Vec(serlaplace(exp(x/(1-5*x)))) \\ _Michel Marcus_, Nov 25 2018

%Y Cf. A000262, A025168, A321837, A321847, A321849, A321850.

%K nonn

%O 0,3

%A _Ludovic Schwob_, Nov 19 2018