A214047 Least m>0 such that n! <= (3/2)^m.
1, 2, 5, 8, 12, 17, 22, 27, 32, 38, 44, 50, 56, 63, 69, 76, 83, 90, 98, 105, 112, 120, 128, 136, 144, 152, 160, 168, 176, 185, 193, 202, 210, 219, 228, 237, 245, 254, 263, 273, 282, 291, 300, 310, 319, 328, 338, 347, 357, 367
Offset: 1
Examples
a(6) = 17 because (3/2)^16 < 6! <= (3/2)^17.
Links
- Clark Kimberling, Table of n, a(n) for n = 1..1000
Programs
-
Mathematica
Table[m=1; While[n!>(3/2)^m, m++]; m, {n,1,100}] Join[{1},With[{c=Log[3/2]},Table[Ceiling[Log[n!]/c],{n,2,50}]]] (* Harvey P. Dale, May 15 2013 *)