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 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). 1
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 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) 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) LINKS Danny Rorabaugh, Table of n, a(n) for n = 1..10000 EXAMPLE 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. PROG (Sage) [x for sublist in [[k for k in [0..ceil(log(factorial(n+1), base=n))] if (factorial(n)

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Last modified August 23 01:10 EDT 2019. Contains 326211 sequences. (Running on oeis4.)