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A269225
Smallest k such that k! > 2^n.
1
2, 3, 3, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 19, 20, 20, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 23, 23, 23, 24, 24, 24, 24, 24, 25, 25, 25, 25, 26, 26, 26, 26, 26, 27, 27, 27, 27
OFFSET
0,1
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
EXAMPLE
a(7) = 6 because 6! = 720 > 2^7 = 128, but 5! = 120 < 128.
MATHEMATICA
a[n_] := Block[{v=2^n, k=1}, While[++k! <= v]; k]; Array[a, 93, 0] (* Giovanni Resta, Jul 11 2016 *)
Module[{nn=30, f}, f=Table[{k, k!}, {k, nn}]; Table[SelectFirst[f, #[[2]]>2^n&], {n, 0, 100}]][[;; , 1]] (* Harvey P. Dale, Feb 19 2024 *)
PROG
(Python)
def a269225():
...k = 1
...f = 1
...p = 1
...n = 0
...while True:
......while f<=p:
.........k += 1
.........f *= k
......yield k
......p *= 2
......n += 1
(PARI) a(n)=localprec(19); my(t=log(2)*n, x=ceil(solve(k=1, n/2+5, lngamma(k+1)-t))); while(x!<=2^n, x++); x \\ Charles R Greathouse IV, Jul 12 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Christian Perfect, Jul 11 2016
STATUS
approved