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A073701
a(n) = n^2*a(n-1)+(-1)^n.
17
1, 0, 1, 8, 129, 3224, 116065, 5687184, 363979777, 29482361936, 2948236193601, 356736579425720, 51370067437303681, 8681541396904322088, 1701582113793247129249, 382855975603480604081024
OFFSET
0,4
COMMENTS
The sequence b(n) := n!^2 satisfies the same recurrence below for a(n) with the initial conditions b(0) = 1, b(1) = 1. It follows that, for n >=3, a(n) = n!^2*(1/(4 + 4/(8 + 9/(15 +...+ (n-1)^2/(n^2-1))))). Hence BesselJ(0,2) := sum {k = 0..inf} (-1)^k/k!^2 = 1/(4 + 4/(8 + 9/(15 + ...+(n-1)^2/(n^2+1 + ...)))) = 0.22388 90779 ... . Cf. A006040. - Peter Bala, Jul 09 2008
LINKS
FORMULA
a(n) = n!^2*Sum_{k=0..n} (-1)^k/k!^2. BesselJ(0, 2*sqrt(x))/(1-x) = Sum_{n>=0} a(n)*x^n/n!^2. a(n) = round(n!^2*BesselJ(0, 2)), n>0.
Recurrence: a(0) = 1, a(1) = 0, a(n) = (n^2-1)*a(n-1) + (n-1)^2*a(n-2), n >= 2. - Peter Bala, Jul 09 2008, corrected by Georg Fischer, Feb 13 2020
MATHEMATICA
Join[{a = 1}, Table[a = a*n^2 + (-1)^n, {n, 15}]] (* Jayanta Basu, Jul 08 2013 *)
PROG
(Magma) [1] cat [ n eq 1 select 0 else n^2*Self(n-1)+(-1)^n:n in [1..15]]; // Marius A. Burtea, Feb 13 2020
CROSSREFS
Sequence in context: A356914 A364986 A338328 * A239756 A295240 A240630
KEYWORD
nonn
AUTHOR
Vladeta Jovovic, Aug 30 2002
STATUS
approved