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
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
KEYWORD
nonn
AUTHOR
Labos Elemer, Jul 21 2000
EXTENSIONS
More terms from Amiram Eldar, Apr 23 2021
STATUS
approved