OFFSET
0,3
COMMENTS
It appears that a(n+1)=2*a(n) if n is in A068499. - Benoit Cloitre, Sep 07 2002
Because a(0) = 1 and for all n > 0, 2*a(n) >= a(n+1), the sequence is a complete sequence. - Frank M Jackson, Aug 09 2013
Luca and Young prove that a(n) divides n! for n >= 6. - Michel Marcus, Nov 02 2017
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)
Daniel Berend and J. E. Harmse, Gaps between consecutive divisors of factorials, Ann. Inst. Fourier, 43 (3) (1993), 569-583.
Paul Erdős, S. W. Graham, Alexsandr Ivić, and Carl Pomerance, On the number of divisors of n!, Analytic Number Theory, Proceedings of a Conference in Honor of Heini Halberstam, ed. by B. C. Berndt, H. G. Diamond, A. J. Hildebrand, Birkhauser 1996, pp. 337-355.
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.
John D. Mahony, The next number in the sequence, Math. Gaz. (2024) Vol. 108, Issue 573, 399-406.
Wikipedia, Complete sequence.
FORMULA
a(n) <= a(n+1) <= 2*a(n) - Benoit Cloitre, Sep 07 2002
From Avik Roy (avik_3.1416(AT)yahoo.co.in), Jan 28 2009: (Start)
Assume, p1,p2...pm are the prime numbers less than or equal to n.
Then, a(n) = Product_{i=1..m} (bi+1), where bk = Sum_{i=1..m} floor(n/pk^i).
For example, if n=5, p1=2,p2=3,p3=5;
b1=floor(5/2)+floor(5/2^2)+floor(5/2^3)+...=2+1+0+..=3 similarly, b2=b3=1;
Thus a(5)=(3+1)(1+1)(1+1)=16. (End)
a(n) ~ exp(c * n/log(n) + O(n/log(n)^2)), where c = A131688 (Erdős et al., 1996). - Amiram Eldar, Nov 07 2020
EXAMPLE
a(4) = 8 because 4!=24 has precisely eight distinct divisors: 1, 2, 3, 4, 6, 8, 12, 24.
MAPLE
A027423 := n -> numtheory[tau](n!);
MATHEMATICA
Table[ DivisorSigma[0, n! ], {n, 0, 35}]
PROG
(PARI) for(k=0, 50, print1(numdiv(k!), ", ")) \\ Jaume Oliver Lafont, Mar 09 2009
(PARI) a(n)=my(s=1, t, tt); forprime(p=2, n, t=tt=n\p; while(tt, t+=tt\=p); s*=t+1); s \\ Charles R Greathouse IV, Feb 08 2013
(Haskell)
a027423 n = f 1 $ map (\p -> iterate (* p) p) a000040_list where
f y ((pps@(p:_)):ppss)
| p <= n = f (y * (sum (map (div n) $ takeWhile (<= n) pps) + 1)) ppss
| otherwise = y
-- Reinhard Zumkeller, Feb 27 2013
(Python 3.8+)
from math import prod
from collections import Counter
from sympy import factorint
def A027423(n): return prod(e+1 for e in sum((Counter(factorint(i)) for i in range(2, n+1)), start=Counter()).values()) # Chai Wah Wu, Jun 25 2022
CROSSREFS
KEYWORD
nonn,easy,nice
AUTHOR
Glen Burch (gburch(AT)erols.com), Leroy Quet.
STATUS
approved