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%I #25 Dec 11 2022 12:14:37
%S 1,1,2,3,8,24,84,420,2268,13440,73920,604800,3931200,33633600,
%T 324324000,3891888000,33081048000,435891456000,4140968832000,
%U 59281238016000,840311548876800,11708340914350080,134645920515025920,2554547108585472000,45616912653312000000
%N a(n) = ceiling(n!/d(n!)).
%C The ceiling function is required only for n = 3 and 5.
%C Luca and Yound prove that a(n) divides n! for n >= 6. - _Michel Marcus_, Nov 02 2017
%C 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
%D 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).
%H Amiram Eldar, <a href="/A055981/b055981.txt">Table of n, a(n) for n = 1..471</a>
%H Florian Luca and Paul Thomas Young, <a href="https://web.math.pmf.unizg.hr/glasnik/vol_47/no2_05.html">On the number of divisors of n! and of the Fibonacci numbers</a>, Glasnik Matematicki, Vol. 47, No. 2 (2012), 285-293. DOI: 10.3336/gm.47.2.05.
%F a(n) = ceiling(A000142(n)/A027423(n)).
%F Sum_{n>=1} 1/a(n) = A071815 - 7/40. - _Amiram Eldar_, Apr 23 2021
%e 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.
%t a[n_] := Ceiling[n!/DivisorSigma[0, n!]]; Array[a, 30] (* _Amiram Eldar_, Apr 23 2021 *)
%o (PARI) a(n) = ceil(n!/numdiv(n!)); \\ _Michel Marcus_, Nov 02 2017
%Y Cf. A000142, A000005, A033950, A027423, A071815.
%K nonn
%O 1,3
%A _Labos Elemer_, Jul 21 2000
%E More terms from _Amiram Eldar_, Apr 23 2021
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