A027423 Number of divisors of n!.
1, 1, 2, 4, 8, 16, 30, 60, 96, 160, 270, 540, 792, 1584, 2592, 4032, 5376, 10752, 14688, 29376, 41040, 60800, 96000, 192000, 242880, 340032, 532224, 677376, 917280, 1834560, 2332800, 4665600, 5529600, 7864320, 12165120, 16422912
Offset: 0
Examples
a(4) = 8 because 4!=24 has precisely eight distinct divisors: 1, 2, 3, 4, 6, 8, 12, 24.
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.
- Index entries for sequences related to factorial numbers
- Index entries for sequences related to divisors of numbers
Programs
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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
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Maple
A027423 := n -> numtheory[tau](n!);
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Mathematica
Table[ DivisorSigma[0, n! ], {n, 0, 35}]
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PARI
for(k=0,50,print1(numdiv(k!),", ")) \\ Jaume Oliver Lafont, Mar 09 2009
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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
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
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