A136432 -id:A136432 - cp's OEIS frontend

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

Alex Ratushnyak, Oct 15 2013

nonn

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

			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.

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.

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

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

a(n) = my(k=1); while (k^n >= k!, k++); k; \\ Michel Marcus, Jan 27 2021

import math
for n in range(333):
  for k in range(1, 100000):
    if math.factorial(k) > k**n:
      print(str(k), end=', ')
      break

A230282 Largest k such that (k*n)! >= (k!)^(n+1).

Original entry on oeis.org

1, 1, 6, 64, 679, 8468, 126784, 2238565, 45605124, 1053117974, 27182818156, 775557529509, 24236473829015, 823299898542083, 30205566231626957, 1190319005015526817, 50143449209799256306, 2248672171655330927835

Offset: 0

Alex Ratushnyak, Oct 14 2013

nonn

			Biggest k such that (3*k)! >= k!^4 is k = 64, so a(3) = 64.
a(10) = 27182818156 because k = 27182818156 satisfies the inequality (k*10)! >= (k!)^11, but k = 27182818157 does not. To verify this, note that taking the logarithm of each side of the inequality gives log((k*10)!) >= 11*log(k!), and use the series expression log(m!) = log(2*Pi*m)/2 + m*log(m) - m + (1/12)/m - (1/360)/m^3 + (1/1260)/m^5 - ... (where the numerators and denominators of the fractions 1/12, -1/360, 1/1260, etc., are from A046968 and A046969, respectively), to get, at k = 27182818156, log(271828181560!) = 6884982704601.26... for the left hand side of the inequality, and the slightly smaller result 11*log(27182818156!) = 6884982704600.83... for the right hand side; then repeat the calculations using k = 27182818157, and observe that this makes the right hand side slightly larger than the left hand side. - _Jon E. Schoenfield_, Oct 23 2013

Cf. A000142, A065027, A136432, A230319.

Table[k = 0; While[(k n)! >= (k!)^(n + 1), k++]; k - 1, {n, 0, 4}] (* T. D. Noe, Oct 18 2013 *)

import math
for n in range(8):
  for k in range(10000000):
    if math.factorial(n*k) < math.factorial(k)**(n+1):
      print(k-1, end=', ')
      break

For n > 1, a(n) = floor(e*(n^n) - ((n^2-1)*log(n) + n*(1+log(2*Pi)))/2) [conjectural, but verified for all n in 2..5000]. - Jon E. Schoenfield, Oct 22 2013

a(7)-a(17) from Jon E. Schoenfield, Oct 22 2013

cp's OEIS Frontend

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

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