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A154029
List of pairs of numbers: {n^2-1, (2*n-1)!!} such that F((2*n-1)!!) = n^2 - 1.
1
0, 1, 3, 3, 8, 15, 15, 105, 24, 945, 35, 10395, 48, 135135, 63, 2027025, 80, 34459425, 99, 654729075, 120, 13749310575, 143, 316234143225, 168, 7905853580625, 195, 213458046676875, 224, 6190283353629375, 255, 191898783962510625, 288
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
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
Sequence in context: A126073 A126592 A055057 * A219349 A208964 A104864
KEYWORD
nonn,tabf
AUTHOR
Roger L. Bagula, Jan 04 2009
STATUS
approved