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A095000 E.g.f.: exp(x)/(1-x)^4. 13
1, 5, 29, 193, 1457, 12341, 116125, 1203329, 13627073, 167525317, 2222710781, 31665408545, 482196718129, 7817359305653, 134443910166077, 2444991262876321, 46883166605035265, 945426638499719429, 20002372214708227933, 443036881445294292737, 10252840082607606694961 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Sum_{k=0..n} A094816(n,k)*x^k give A000522(n), A001339(n), A082030(n) for x = 1, 2, 3 respectively.
From Peter Bala, Jul 10 2008: (Start)
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.
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))))).
Lim_{n -> infinity} a(n)/b(n) = e/6 = 1/(2+1/(5-1/(6-2/(7-...-n/((n+5)-...))))).
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)).
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).
{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)
LINKS
Eric Weisstein's World of Mathematics, Poisson-Charlier polynomial
FORMULA
a(n) = Sum_{k=0..n} A094816(n, k)*4^k.
a(n) = Sum_{k=0..n} binomial(n, k)*(k+3)!/6.
a(n) ~ n!*n^3*e/6. - Vaclav Kotesovec, Oct 14 2012
a(n) = hypergeom([4, -n], [], -1). - Peter Luschny, Sep 20 2014
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
D-finite with recurrence a(n) +(-n-4)*a(n-1) +(n-1)*a(n-2)=0. - R. J. Mathar, Aug 01 2022
MAPLE
a := n -> hypergeom([4, -n], [], -1); seq(round(evalf(a(n), 100)), n=0..18); # Peter Luschny, Sep 20 2014
MATHEMATICA
Table[n!*SeriesCoefficient[E^(x)/(1-x)^4, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 14 2012 *)
PROG
(PARI) x='x+O('x^66); Vec(serlaplace(exp(x)/(1-x)^4)) \\ Joerg Arndt, May 11 2013
CROSSREFS
Sequence in context: A258314 A225030 A188143 * A086672 A324962 A306932
KEYWORD
nonn
AUTHOR
Philippe Deléham, Jun 19 2004
STATUS
approved

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