A055981 a(n) = ceiling(n!/d(n!)).

1, 1, 2, 3, 8, 24, 84, 420, 2268, 13440, 73920, 604800, 3931200, 33633600, 324324000, 3891888000, 33081048000, 435891456000, 4140968832000, 59281238016000, 840311548876800, 11708340914350080, 134645920515025920, 2554547108585472000, 45616912653312000000

Offset

1,3

Comments

The ceiling function is required only for n = 3 and 5.

Luca and Yound prove that a(n) divides n! for n >= 6. - Michel Marcus, Nov 02 2017

Problem 3 in the 1976 Miklós Schweitzer Competition is to show that tau(n!) divides n! for all sufficiently large n. - Martin Renner, Dec 09 2022

References

Gábor J. Székely (ed.), Contests in Higher Mathematics. Miklós Schweitzer Competitions 1962-1991. With 39 illustrations. New York: Springer, 1996. (Problem Books in Mathematics.), p. 23 (problem 1976, nr. 3), 376-378 (solution).

Links

Amiram Eldar, Table of n, a(n) for n = 1..471

Florian Luca and Paul Thomas Young, On the number of divisors of n! and of the Fibonacci numbers, Glasnik Matematicki, Vol. 47, No. 2 (2012), 285-293. DOI: 10.3336/gm.47.2.05.

Formula

a(n) = ceiling(A000142(n)/A027423(n)).

Sum_{n>=1} 1/a(n) = A071815 - 7/40. - Amiram Eldar, Apr 23 2021

Example

For n=3 n!=6, d(n!)=4, quotient is 3/2, for n=5 n!=120, d(n!)=16, quotient=15/2. All other cases give integers.

Mathematica

a[n_] := Ceiling[n!/DivisorSigma[0, n!]]; Array[a, 30] (* Amiram Eldar, Apr 23 2021 *)

Prog

(PARI) a(n) = ceil(n!/numdiv(n!)); \\ Michel Marcus, Nov 02 2017

Crossrefs

Cf. A000142, A000005, A033950, A027423, A071815.

Sequence in context: A089847 A304121 A038561 * A182212 A120260 A202592

Adjacent sequences: A055978 A055979 A055980 * A055982 A055983 A055984

Keyword

nonn

Author

Labos Elemer, Jul 21 2000

Extensions

More terms from Amiram Eldar, Apr 23 2021

Status

approved