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

%I A230319 #41 Nov 12 2024 18:59:58
%S A230319 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,
%T A230319 34,36,37,38,39,41,42,43,44,46,47,48,49,51,52,53,54,55,57,58,59,60,62,
%U A230319 63,64,65,67,68,69,70,71,73,74,75,76,78,79,80,81,82,84,85,86,87,88
%N A230319 Least positive k such that k! > k^n.
%C A230319 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, ...
%C A230319 It appears that a(n) = A277675(n) + 2 for n >= 1. - _Hugo Pfoertner_, Jan 27 2021
%C A230319 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
%H A230319 Jon E. Schoenfield, <a href="/A230319/b230319.txt">Table of n, a(n) for n = 0..10000</a>
%H A230319 M. Sánchez Garza and E. Treviño, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL24/Trevino/trevino6.html">On a sequence related to the factoradic representation of an integer</a>, Journal of Integer Sequences Vol. 24 (2021), Article 21.8.5.
%e A230319 Least k>0 such that k! > k^3 is k=6.
%e A230319 For k=5: 5! = 120 < 125 = 5^3.
%e A230319 For k=6: 6! = 720 > 216 = 6^3.
%e A230319 So a(3) = 6.
%t A230319 Table[k = 1; While[k^n >= k!, k++]; k, {n, 0, 100}] (* _T. D. Noe_, Oct 18 2013 *)
%o A230319 (Python)
%o A230319 import math
%o A230319 for n in range(333):
%o A230319   for k in range(1, 100000):
%o A230319     if math.factorial(k) > k**n:
%o A230319       print(str(k), end=', ')
%o A230319       break
%o A230319 (PARI) a(n) = my(k=1); while (k^n >= k!, k++); k; \\ _Michel Marcus_, Jan 27 2021
%Y A230319 Cf. A000142, A065027, A136432, A230282, A336803.
%K A230319 nonn
%O A230319 0,1
%A A230319 _Alex Ratushnyak_, Oct 15 2013