A123022 a(n) = n!*b(n) where b(n) = (n-4)*b(n-2)/(n*(n-1)) and b(0) = b(1) = 1.
1, 1, -2, -1, 0, -1, 0, -3, 0, -15, 0, -105, 0, -945, 0, -10395, 0, -135135, 0, -2027025, 0, -34459425, 0, -654729075, 0, -13749310575, 0, -316234143225, 0, -7905853580625, 0, -213458046676875, 0, -6190283353629375, 0, -191898783962510625, 0, -6332659870762850625, 0
Offset: 0
Keywords
References
- Richard Bronson, Schaum's Outline of Modern Introductory Differential Equations, MacGraw-Hill, New York,1973, page 99, solved problem 19.1.
Links
- G. C. Greubel, Table of n, a(n) for n = 0..750
Crossrefs
Cf. A330797.
Programs
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Magma
[n le 2 select 1 else (n-5)*Self(n-2): n in [1..30]]; // G. C. Greubel, Jul 11 2021
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Maple
b[0]:=1: b[1]:=1: for n from 2 to 40 do b[n]:=(n-4)*b[n-2]/(n*(n-1)) od: seq(n!*b[n], n=0..40);
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Mathematica
b[n_]:= b[n]= If[n<2, 1, (n-4)*b[n-2]/(n*(n-1))]; a[n_]:= n!*b[n]; Table[a[n], {n, 0, 30}] -
Sage
def a(n): return 1 if (n<2) else (n-4)*a(n-2) [a(n) for n in (0..30)] # G. C. Greubel, Jul 11 2021
Formula
G.f.: (1 - G(0))*x^4/(1+x) +1 +x -2*x^2 -x^3, where G(k)= 1 + x*(2*k+1)/(1 - x/(x + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 07 2013
From G. C. Greubel, Jul 10 2021: (Start)
a(2*n) = (-2)^n*[n<2], for n >= 0.
a(2*n+1) = 2^n*Pochhammer(n, 1/2)/(1-2*n), for n >= 0.
a(2*n+1) = (-1)^n * A330797(n).
D-finite with recurrence a(n) = (n-4)*a(n-2) with a(0) = a(1) = 1.
G.f.: 1 + x - 2*x^2 - sqrt(Pi/2)*x^2*exp(-1/(2*x^2))*erfi(1/(sqrt(2)*x)). (End)
Extensions
Edited by N. J. A. Sloane, Oct 01 2006 and Nov 24 2006