%I #39 Aug 01 2022 09:30:58
%S 1,5,29,193,1457,12341,116125,1203329,13627073,167525317,2222710781,
%T 31665408545,482196718129,7817359305653,134443910166077,
%U 2444991262876321,46883166605035265,945426638499719429,20002372214708227933,443036881445294292737,10252840082607606694961
%N E.g.f.: exp(x)/(1-x)^4.
%C Sum_{k=0..n} A094816(n,k)*x^k give A000522(n), A001339(n), A082030(n) for x = 1, 2, 3 respectively.
%C From _Peter Bala_, Jul 10 2008: (Start)
%C Recurrence relation: a(0) = 1, a(1) = 5, a(n) = (n+4)*a(n-1) - (n-1)*a(n-2) for n >= 2. Let p_3(n) = n^3+2*n-1 = n^(3)-3*n^(2)+3*n^(1)-1, where n^(k) denotes the rising factorial n*(n+1)*...*(n+k-1). The polynomial p_3(n) is an example of a Poisson-Charlier polynomial c_k(x;a) at k = 3, x = -n and a = -1.
%C The sequence b(n) := n!*p_3(n+1) = A001565(n) satisfies the same recurrence as a(n) but with the initial conditions b(0) = 2, b(1) = 11. This leads to the finite continued fraction expansion a(n)/b(n) = 1/(2+1/(5-1/(6-2/(7-...-(n-1)/(n+4))))).
%C Lim_{n -> infinity} a(n)/b(n) = e/6 = 1/(2+1/(5-1/(6-2/(7-...-n/((n+5)-...))))).
%C a(n) = -b(n) * Sum_{k = 0..n} 1/(k!*p_3(k)*p_3(k+1)) - since the rhs satisfies the above recurrence with the same initial conditions. Hence e = -6 * Sum_{k>=0} 1/(k!*p_3(k)*p_3(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) and A095177 (r=4).
%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. (End)
%H Vincenzo Librandi, <a href="/A095000/b095000.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)*4^k.
%F a(n) = Sum_{k=0..n} binomial(n, k)*(k+3)!/6.
%F a(n) ~ n!*n^3*e/6. - _Vaclav Kotesovec_, Oct 14 2012
%F a(n) = hypergeom([4, -n], [], -1). - _Peter Luschny_, Sep 20 2014
%F First-order recurrence: P(n-1)*a(n) = n*P(n)*a(n-1) - 1 with a(0) = 1, where P(n) = n^3 + 3*n^2 + 5*n + 2 = A001565(n). - _Peter Bala_, Jul 26 2021
%F D-finite with recurrence a(n) +(-n-4)*a(n-1) +(n-1)*a(n-2)=0. - _R. J. Mathar_, Aug 01 2022
%p a := n -> hypergeom([4, -n], [], -1); seq(round(evalf(a(n), 100)), n=0..18); # _Peter Luschny_, Sep 20 2014
%t Table[n!*SeriesCoefficient[E^(x)/(1-x)^4,{x,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 14 2012 *)
%o (PARI) x='x+O('x^66); Vec(serlaplace(exp(x)/(1-x)^4)) \\ _Joerg Arndt_, May 11 2013
%Y Cf. A000522, A001339, A082030, A095177, A096307, A096341, A001565.
%K nonn
%O 0,2
%A _Philippe Deléham_, Jun 19 2004