Juan Castaneda - cp's OEIS frontend

User: Juan Castaneda

Juan Castaneda has authored 2 sequences.

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

Juan Castaneda, Mar 04 2015

nonn

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)

			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.

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

Cf. A060151, A074181-A074184, A111683.

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

More terms from Tom Edgar, Mar 04 2015

A256774 All factorials n! along with powers of the numbers n and n+1 that fall in between n! and (n+1)!, in increasing order.

Original entry on oeis.org

1, 2, 3, 4, 6, 9, 16, 24, 25, 64, 120, 125, 216, 625, 720, 1296, 2401, 5040, 16807, 32768, 40320, 59049, 262144, 362880, 531441, 1000000, 3628800, 10000000, 19487171, 39916800, 214358881, 429981696, 479001600, 815730721, 5159780352, 6227020800, 10604499373, 20661046784, 87178291200, 289254654976

Offset: 1

Juan Castaneda, Apr 10 2015

For each positive integer n, we consider the two factorials n! and (n+1)! as lower and upper bounds of an interval. Then we look for all powers of n and all powers of n+1 that fall inside that interval. We sort those numbers in increasing order, and we append them to the sequence without allowing duplicates. Then we move on to the next integer, and so on.

A000142 (without its first term that stands for 0!) is a subsequence.

			With n=1: 1! < 2! gives a(1)=1, a(2)=2.
With n=2: 2! < 3^1 < 2^2 < 3! gives a(3)=3, a(4)=4, a(5)=6.
With n=3: 3! < 3^2 < 4^2 < 4! gives a(6)=9, a(7)=16, a(8)=24.
With n=4: 4! < 5^2 < 4^3 < 5! gives a(9)=25, a(10)=64, a(11)=120.
With n=5: 5! < 5^3 < 6^3 < 5^4 < 6! gives a(12)=125, a(13)=216, a(14)=625, a(15)=720

Michael De Vlieger, Table of n, a(n) for n = 1..1346

Cf. A000142, A039960, A060151, A074181, A074182, A074184, A111683.

f[n_] := Block[{a = n!, b = (n + 1)!}, Sort@ Union[{a}, n^Range[Ceiling@ Log[n, a], Floor@ Log[n, b]], (n + 1)^Range[Ceiling@ Log[n + 1, a], Floor@ Log[n + 1, b]]]]; {1}~Join~(f /@ Range[2, 14] // Flatten) (* Michael De Vlieger, Apr 15 2015 *)

tabf(nn) = {print([1]); for (n=2, nn, v = [n!]; ka = ceil(log(n!+1)/log(n)); kb = floor(log((n+1)!-1)/log(n)); for (k=ka, kb, v = concat(v, n^k);); ka = ceil(log(n!+1)/log(n+1)); kb = floor(log((n+1)!-1)/log(n+1)); for (k=ka, kb, v = concat(v, (n+1)^k);); print(vecsort(v));); } \\ Michel Marcus, Apr 22 2015

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User: Juan Castaneda

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).

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