A230319 Least positive k such that k! > k^n.

2, 3, 4, 6, 7, 8, 10, 11, 12, 14, 15, 16, 18, 19, 20, 22, 23, 24, 25, 27, 28, 29, 31, 32, 33, 34, 36, 37, 38, 39, 41, 42, 43, 44, 46, 47, 48, 49, 51, 52, 53, 54, 55, 57, 58, 59, 60, 62, 63, 64, 65, 67, 68, 69, 70, 71, 73, 74, 75, 76, 78, 79, 80, 81, 82, 84, 85, 86, 87, 88

Offset

0,1

Comments

Numbers that are not in the sequence: 0, 1, 5, 9, 13, 17, 21, 26, 30, 35, 40, 45, 50, 56, 61, 66, 72, 77, 83, 89, 95, 100, 106, 112, 118, 124, 130, 137, 143, 149, 155, 161, 168, ...

It appears that a(n) = A277675(n) + 2 for n >= 1. - Hugo Pfoertner, Jan 27 2021

Sánchez Garza and Treviño proved that the difference between any two consecutive elements is 1 or 2 and that the counting function up to x is x+x/log x + o(x/log x). - Enrique Treviño, Jan 30 2021

Links

Jon E. Schoenfield, Table of n, a(n) for n = 0..10000

M. Sánchez Garza and E. Treviño, On a sequence related to the factoradic representation of an integer, Journal of Integer Sequences Vol. 24 (2021), Article 21.8.5.

Example

Least k>0 such that k! > k^3 is k=6.
For k=5: 5! = 120 < 125 = 5^3.
For k=6: 6! = 720 > 216 = 6^3.
So a(3) = 6.

Mathematica

Table[k = 1; While[k^n >= k!, k++]; k, {n, 0, 100}] (* T. D. Noe, Oct 18 2013 *)

Prog

(Python)
import math
for n in range(333):
  for k in range(1, 100000):
    if math.factorial(k) > k**n:
      print(str(k), end=', ')
      break
(PARI) a(n) = my(k=1); while (k^n >= k!, k++); k; \\ Michel Marcus, Jan 27 2021

Crossrefs

Cf. A000142, A065027, A136432, A230282, A336803.

Sequence in context: A187572 A184478 A188188 * A319371 A004772 A376455

Adjacent sequences: A230316 A230317 A230318 * A230320 A230321 A230322

Keyword

nonn,changed

Author

Alex Ratushnyak, Oct 15 2013

Status

approved