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
Keywords
Examples
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.
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.
Programs
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Mathematica
Table[k = 1; While[k^n >= k!, k++]; k, {n, 0, 100}] (* T. D. Noe, Oct 18 2013 *)
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PARI
a(n) = my(k=1); while (k^n >= k!, k++); k; \\ Michel Marcus, Jan 27 2021
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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
Comments