cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A027423 Number of divisors of n!.

Original entry on oeis.org

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

Views

Author

Glen Burch (gburch(AT)erols.com), Leroy Quet

Keywords

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

Examples

			a(4) = 8 because 4!=24 has precisely eight distinct divisors: 1, 2, 3, 4, 6, 8, 12, 24.
		

Crossrefs

Cf. A000005, A000142, A062569, A131688, A161466 (divisors of 10!).

Programs

  • 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
  • Maple
    A027423 := n -> numtheory[tau](n!);
  • Mathematica
    Table[ DivisorSigma[0, n! ], {n, 0, 35}]
  • 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
    

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) = A000005(A000142(n)). - Michel Marcus, Sep 13 2014
a(n) ~ exp(c * n/log(n) + O(n/log(n)^2)), where c = A131688 (Erdős et al., 1996). - Amiram Eldar, Nov 07 2020