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).

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, 21, 22, 22, 23, 24, 25, 26, 26, 27, 28, 29, 30, 30, 31, 32, 33, 34, 34, 35, 36, 37, 38, 38, 39, 40, 41, 42, 43, 43, 44, 45, 46, 47, 47, 48, 49, 50, 51, 51, 52, 53, 54

Offset: 1

Author

Juan Castaneda, Mar 04 2015

Keywords

• nonn

Comments

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.

The sequence { floor(log_n((n+1)!)) | n>=2 } is a subsequence.

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)Danny Rorabaugh, Apr 03 2015

From Danny Rorabaugh, Apr 15 2015: (Start)

This sequence is the same as A074184 for 6<=n<=10000.

For k > 2, k! < k^(ceiling(log_k(k!))) < (k+1)!.

The two sequences continue to be identical provided k^(1 + ceiling(log_k(k!))) > (k+1)! when k > 5.

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.

Are either of the final inequalities true for all sufficiently large k?

(End)

Examples

			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.
		

Links

• Danny Rorabaugh, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A060151, A074181-A074184, A111683.

Programs

Sage

[x for sublist in [[k for k in [0..ceil(log(factorial(n+1),base=n))] if (factorial(n)Tom Edgar, Mar 04 2015

Extensions

More terms from Tom Edgar, Mar 04 2015