A269225 Smallest k such that k! > 2^n.

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

Author

Christian Perfect, Jul 11 2016

Keywords

• nonn
• easy

Examples

			a(7) = 6 because 6! = 720 > 2^7 = 128, but 5! = 120 < 128.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A003070, A067850, A258782.

Programs

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 *)

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

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