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.

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

Author

Juan Castaneda, Apr 10 2015

Keywords

• nonn
• tabf

Comments

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.

Examples

			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
		

Links

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

Crossrefs

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

Programs

Mathematica

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 *)

PARI

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