|
Search: id:A123761
|
|
|
| A123761 |
|
Let k(n)=mod(3,n)-1. Then a(n) = 4*a(n-1) if n is odd, otherwise ((5+k(n))/4)*a(n-1), with a(0) = 1, a(1) = 2. |
|
+0
1
|
|
| 1, 2, 3, 12, 15, 60, 60, 240, 360, 1440, 1800, 7200, 7200, 28800, 43200, 172800, 216000, 864000, 864000, 3456000, 5184000, 20736000, 25920000, 103680000, 103680000, 414720000, 622080000, 2488320000, 3110400000, 12441600000
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
A double modulo switch recursion with four basic ratio states: {4,1,5/4,3/2}.
Surprisingly, the function behaves very much like the factorial function.
|
|
MATHEMATICA
|
k[n_] := Mod[n, 3] - 1; f[0] = 1; f[1] = 2; f[n_] := f[n] = If[Mod[n, 2] == 1, 4*f[n - 1], ((5 + k[n])/4)*f[n - 1]]; a = Table[f[n], {n, 0, 30}]
|
|
CROSSREFS
|
Sequence in context: A076175 A067780 A124486 this_sequence A047163 A046486 A073452
Adjacent sequences: A123758 A123759 A123760 this_sequence A123762 A123763 A123764
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Roger Bagula (rlbagulatftn(AT)yahoo.com), Nov 16 2006
|
|
EXTENSIONS
|
Edited by njas, Nov 19 2006
|
|
|
Search completed in 0.002 seconds
|
