OFFSET
0,5
COMMENTS
a(n) = n + a(n-1), a(n) = n*(n+1)/2, with a(0) = 0 and b(n) = n*a(n-1), a(n) = n!, with b(0) = 1, then you can form a function F such that: F(n!) = n*(n+1)/2.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..900
MATHEMATICA
Flatten[Table[{n*(n + 1)/2, n!}, {n, 0, 20}]] (* Produces terms *)
(*addition*)
a[0] = 0; a[n_] := a[n] = n + a[n - 1];
Table[a[n] - n*(n + 1)/2, {n, 0, 20}] (* Demonstrates that a[n] = n*(n+1)/2 *)
(*multiplication*)
b[0] = 1; b[n_] := b[n] = n*b[n - 1];
Table[b[n] - n!, {n, 0, 20}] (* Demonstrates that b[n] = n! *)
PROG
(PARI) a(n)=if(n%2, (n\2)!, n*(n+2)/8) \\ Charles R Greathouse IV, Sep 01 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Roger L. Bagula, Jan 04 2009
STATUS
approved