A308852 Minimum number k such that the k-th tetrahedral number is not smaller than n!.

1, 2, 3, 5, 8, 16, 31, 62, 129, 279, 621, 1421, 3343, 8057, 19870, 50071, 128747, 337414, 900358, 2443947, 6742667, 18893218, 53729800, 154983562, 453174686, 1342528227, 4027584682, 12230119228, 37574801086, 116753643340, 366767636286, 1164414663338, 3734900007009

Offset: 1

Author

Lorenzo Sauras Altuzarra, Jun 28 2019

Keywords

• nonn

Comments

More formally, a(n) is the minimum element of the set of positive integers k such that the k-th tetrahedral number is not smaller than the n-th factorial.

Open problem: what is the cardinality of the set of numbers that are simultaneously a tetrahedral number and a factorial number? For example, 1 and 120 belong to this set.

Examples

			The minimum tetrahedral number not smaller than 4! is 35 (i.e., the 5th tetrahedral number), so a(4) = 5.
The minimum tetrahedral number not smaller (equal, in fact) than 5! is 120 (i.e., the 8th tetrahedral number), so a(5) = 8.
		

Crossrefs

Cf. A000142 (factorial numbers), A000292 (tetrahedral numbers).

Cf. A055228 (for n^2), A214049 (for n^3), A214448 (for n^4).

Programs

Mathematica

Floor[(6 Range[33]!)^(1/3)] (* Giovanni Resta, Jul 30 2019 *)

PARI

a(n) = {my(k=1); while (k*(k+1)*(k+2)/6 < n!, k++); k;} \\ Michel Marcus, Jun 28 2019

Formula

a(n) = ceiling((sqrt(3) * sqrt(243*(n!)^2 - 1) + 27*n!)^(1/3) / 3^(2/3) + 1/(3^(1/3) * (sqrt(3) * sqrt(243*(n!)^2 - 1) + 27*n!)^(1/3)) - 1). - Daniel Suteu, Jun 30 2019

a(n) = floor((6*n!)^(1/3)). - Giovanni Resta, Jul 30 2019

Extensions

a(26)-a(33) from Daniel Suteu, Jun 30 2019