OFFSET
0,3
COMMENTS
If you have two recursions ( addition and multiplication):
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.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..800
FORMULA
{n^2 - 1, (2*n - 1)!!}.
MATHEMATICA
Clear[a, b, n];
Flatten[Table[{n^2 - 1, (2*n - 1)!!}, {n, 1, 20}]] (* Produces terms *)
(*addition*)
a[0] = -1; a[n_] := a[n] = (2*n - 1) + a[n - 1];
Table[a[n] - (n^2 - 1), {n, 0, 20}] (* Demonstrates that a[n] = n^2 -1 *)
(*multiplication*)
b[0] = 1; b[n_] := b[n] = (2*n - 1)*b[n - 1];
Table[b[n] - (2*n - 1)!!, {n, 0, 20}] (* Demonstrates that b[n] = (2*n - 1)!! *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Roger L. Bagula, Jan 04 2009
STATUS
approved