cp's OEIS Frontend

A248868 Exponents n that make k! < k^n < (k+1)! hold true for some integer k > 1, in increasing order by k, then n (if applicable).

%I A248868 #35 May 10 2015 09:54:16
%S A248868 2,2,3,3,4,4,5,6,6,7,8,9,9,10,11,12,12,13,14,15,15,16,17,18,19,19,20,
%T A248868 21,22,22,23,24,25,26,26,27,28,29,30,30,31,32,33,34,34,35,36,37,38,38,
%U A248868 39,40,41,42,43,43,44,45,46,47,47,48,49,50,51,51,52,53,54
%N A248868 Exponents n that make k! < k^n < (k+1)! hold true for some integer k > 1, in increasing order by k, then n (if applicable).
%C A248868 This sequence consists of those positive integers that, when taken as exponents of some positive integer greater than 1, make the corresponding power of that other integer fall strictly between its factorial and the factorial of the next integer, as shown in the examples.
%C A248868 The sequence { floor(log_n((n+1)!)) | n>=2 } is a subsequence.
%C A248868 This sequence is nondecreasing. Indeed for k>1, k^n<(k+1)! implies n<=k, which implies ((k+1)/k)^(n-1) <= (1 + 1/k)^(k-1) = Sum_{i=0..k-1} binomial(k-1,i) (1/k)^i < Sum_{i=0..k-1} ((k-1)/k)^i < k, which implies (k+1)^(n-1)<k^n<(k+1)!. - _Danny Rorabaugh_, Apr 03 2015
%C A248868 From _Danny Rorabaugh_, Apr 15 2015: (Start)
%C A248868 This sequence is the same as A074184 for 6<=n<=10000.
%C A248868 For k > 2, k! < k^(ceiling(log_k(k!))) < (k+1)!.
%C A248868 The two sequences continue to be identical provided k^(1 + ceiling(log_k(k!))) > (k+1)! when k > 5.
%C A248868 This is equivalent to k^(2 - fractional_part(log_k(k!))) > k + 1, which can be approximated by fractional_part(1/2 - (k + sqrt(2*Pi))/log(k)) < 1 - 1/(k*log(k)) using Stirling's approximation.
%C A248868 Are either of the final inequalities true for all sufficiently large k?
%C A248868 (End)
%H A248868 Danny Rorabaugh, <a href="/A248868/b248868.txt">Table of n, a(n) for n = 1..10000</a>
%e A248868 2! < 2^2 < 3! < 3^2 < 4! < 4^3 < 5! < 5^3 < 5^4 < 6! < 6^4 < 7! < 7^5 < 8! and so on; this sequence consists of the exponents.
%o A248868 (Sage)
%o A248868 [x for sublist in [[k for k in [0..ceil(log(factorial(n+1),base=n))] if (factorial(n)<n^k and n^k<factorial(n+1))] for n in [2..100]] for x in sublist] # _Tom Edgar_, Mar 04 2015
%Y A248868 Cf. A060151, A074181-A074184, A111683.
%K A248868 nonn
%O A248868 1,1
%A A248868 _Juan Castaneda_, Mar 04 2015
%E A248868 More terms from _Tom Edgar_, Mar 04 2015