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%I #29 Jul 27 2021 21:21:07
%S 1,6,41,316,2721,25946,271801,3105936,38474561,513796366,7360674441,
%T 112632827396,1833790646881,31656637715106,577636838177561,
%U 11109543835539736,224635867973671041,4764236394052127126
%N E.g.f.: exp(x)/(1-x)^5.
%C Sum_{k = 0..n} A094816(n,k)*x^k give A000522(n), A001339(n), A082030(n), A095000(n) for x = 1, 2, 3, 4 respectively.
%C From _Peter Bala_, Jul 10 2008: (Start)
%C a(n) is a difference divisibility sequence, that is, the difference a(n) - a(m) is divisible by n - m for all n and m (provided n is not equal to m). See A000522 for further properties of difference divisibility sequences.
%C Recurrence relation: a(0) = 1, a(1) = 6, a(n) = (n+5)*a(n-1) - (n-1)*a(n-2) for n >= 2. Let p_4(n) = n^4+2*n^3+5*n^2+1 = n^(4)-4*n^(3)+6*n^(2)-4*n^(1)+1, where n^(k) denotes the rising factorial n*(n+1)*...*(n+k-1). The polynomial p_4(n) is an example of a Poisson-Charlier polynomial c_k(x;a) at k = 4, x = -n and a = -1.
%C The sequence b(n) := n!*p_4(n+1) = A001688(n) satisfies the same recurrence as a(n) but with the initial conditions b(0) = 9, b(1) = 53. This leads to the finite continued fraction expansion expansion a(n)/b(n) = 1/(9-1/(6-1/(7-2/(8-...-(n-1)/(n+5))))).
%C Lim n -> infinity a(n)/b(n) = e/24 = 1/(9-1/(6-1/(7-2/(8-...-n/((n+6)-...))))).
%C a(n) = b(n) * sum {k = 0..n} 1/(k!*p_4(k)*p_4(k+1)) - since the rhs satisfies the above recurrence with the same initial conditions. Hence e = 24 * sum {k = 0..inf} 1/(k!*p_4(k)p_4(k+1)).
%C For sequences satisfying the more general recurrence a(n) = (n+1+r)*a(n-1) - (n-1)*a(n-2), which yield series acceleration formulas for e/r! that involve the Poisson-Charlier polynomials c_r(-n;-1), refer to A000522 (r = 0), A001339 (r=1), A082030 (r=2), A095000 (r=3). (End)
%H Vincenzo Librandi, <a href="/A095177/b095177.txt">Table of n, a(n) for n = 0..200</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Poisson-CharlierPolynomial.html">Poisson-Charlier polynomial</a>
%F a(n) = Sum_{k = 0..n} A094816(n, k)*5^k.
%F a(n) = Sum_{k=0..n} binomial(n, k)*(k+4)!/4!.
%F G.f.: 1/Q(0), where Q(k) = 1 - x - x*(k+5)/(1 - x*(k+1)/Q(k+1)); (continued fraction). - _Sergei N. Gladkovskii_, Apr 22 2013
%F a(n) ~ n! *exp(1)*n^4/24. - _Vaclav Kotesovec_, Jun 21 2013
%F a(n) = 2F0(5,-n;;-1). - _Benedict W. J. Irwin_, May 27 2016
%F First-order recurrence: P(n-1)*a(n) = n*P(n)*a(n-1) + 1 with a(0) = 1, where P(n) = n^4 + 6*n^3 + 17*n^2 + 20*n + 9 = A094793(n). - _Peter Bala_, Jul 26 2021
%t CoefficientList[Series[Exp[x]/(1-x)^5, {x, 0, 20}], x]* Range[0, 20]! (* _Vaclav Kotesovec_, Jun 21 2013 *)
%t Table[HypergeometricPFQ[{5, -n}, {}, -1], {n, 0, 20}] (* _Benedict W. J. Irwin_, May 27 2016 *)
%o (PARI) a(n) = sum(k=0,n, binomial(n, k)*(k+4)!/4! ); \\ _Joerg Arndt_, Apr 22 2013
%Y Cf. A000522, A001339, A082030, A095000, A096307, A096341, A094793.
%K nonn
%O 0,2
%A _Philippe Deléham_, Jun 20 2004
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