|
%I #7 Sep 03 2016 17:06:04
%S 0,1,3,3,8,15,15,105,24,945,35,10395,48,135135,63,2027025,80,34459425,
%T 99,654729075,120,13749310575,143,316234143225,168,7905853580625,195,
%U 213458046676875,224,6190283353629375,255,191898783962510625,288
%N List of pairs of numbers: {n^2-1, (2*n-1)!!} such that F((2*n-1)!!) = n^2 - 1.
%C If you have two recursions ( addition and multiplication):
%C a(n) = 2*n - 1 + a(n-1), a(n) = n*(n+1)/2, with a(0) = -1 and b(n) = (2*n - 1)*a(n-1), a(n) = n!, with b(0) = 1 then you can form a function F such that: F((2*n-1)!!) = n^2 - 1.
%H G. C. Greubel, <a href="/A154029/b154029.txt">Table of n, a(n) for n = 0..800</a>
%F {n^2 - 1, (2*n - 1)!!}.
%t Clear[a,b,n];
%t Flatten[Table[{n^2 - 1, (2*n - 1)!!}, {n,1,20}]] (* Produces terms *)
%t (*addition*)
%t a[0] = -1; a[n_] := a[n] = (2*n - 1) + a[n - 1];
%t Table[a[n] - (n^2 - 1), {n, 0, 20}] (* Demonstrates that a[n] = n^2 -1 *)
%t (*multiplication*)
%t b[0] = 1; b[n_] := b[n] = (2*n - 1)*b[n - 1];
%t Table[b[n] - (2*n - 1)!!, {n, 0, 20}] (* Demonstrates that b[n] = (2*n - 1)!! *)
%K nonn,tabf
%O 0,3
%A _Roger L. Bagula_, Jan 04 2009
|
