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A123022 a(n) = n!*b(n) where b(n) = (n-4)*b(n-2)/(n*(n-1)) and b(0) = b(1) = 1.

%I A123022 #23 Sep 08 2022 08:45:28
%S A123022 1,1,-2,-1,0,-1,0,-3,0,-15,0,-105,0,-945,0,-10395,0,-135135,0,
%T A123022 -2027025,0,-34459425,0,-654729075,0,-13749310575,0,-316234143225,0,
%U A123022 -7905853580625,0,-213458046676875,0,-6190283353629375,0,-191898783962510625,0,-6332659870762850625,0
%N A123022 a(n) = n!*b(n) where b(n) = (n-4)*b(n-2)/(n*(n-1)) and b(0) = b(1) = 1.
%D A123022 Richard Bronson, Schaum's Outline of Modern Introductory Differential Equations, MacGraw-Hill, New York,1973, page 99, solved problem 19.1.
%H A123022 G. C. Greubel, <a href="/A123022/b123022.txt">Table of n, a(n) for n = 0..750</a>
%F A123022 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
%F A123022 From _G. C. Greubel_, Jul 10 2021: (Start)
%F A123022 a(2*n) = (-2)^n*[n<2], for n >= 0.
%F A123022 a(2*n+1) = 2^n*Pochhammer(n, 1/2)/(1-2*n), for n >= 0.
%F A123022 a(2*n+1) = (-1)^n * A330797(n).
%F A123022 D-finite with recurrence a(n) = (n-4)*a(n-2) with a(0) = a(1) = 1.
%F A123022 G.f.: 1 + x - 2*x^2 - sqrt(Pi/2)*x^2*exp(-1/(2*x^2))*erfi(1/(sqrt(2)*x)). (End)
%p A123022 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);
%t A123022 b[n_]:= b[n]= If[n<2, 1, (n-4)*b[n-2]/(n*(n-1))]; a[n_]:= n!*b[n];
%t A123022 Table[a[n], {n, 0, 30}]
%o A123022 (Magma) [n le 2 select 1 else (n-5)*Self(n-2): n in [1..30]]; // _G. C. Greubel_, Jul 11 2021
%o A123022 (Sage)
%o A123022 def a(n): return 1 if (n<2) else (n-4)*a(n-2)
%o A123022 [a(n) for n in (0..30)] # _G. C. Greubel_, Jul 11 2021
%Y A123022 Cf. A330797.
%K A123022 sign
%O A123022 0,3
%A A123022 _Roger L. Bagula_, Sep 24 2006
%E A123022 Edited by _N. J. A. Sloane_, Oct 01 2006 and Nov 24 2006