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%I #74 May 20 2023 01:17:50
%S 1,1,2,4,8,16,30,60,96,160,270,540,792,1584,2592,4032,5376,10752,
%T 14688,29376,41040,60800,96000,192000,242880,340032,532224,677376,
%U 917280,1834560,2332800,4665600,5529600,7864320,12165120,16422912
%N Number of divisors of n!.
%C It appears that a(n+1)=2*a(n) if n is in A068499. - _Benoit Cloitre_, Sep 07 2002
%C 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
%C Luca and Young prove that a(n) divides n! for n >= 6. - _Michel Marcus_, Nov 02 2017
%H Seiichi Manyama, <a href="/A027423/b027423.txt">Table of n, a(n) for n = 0..10000</a> (terms 0..1000 from T. D. Noe)
%H Daniel Berend and J. E. Harmse, <a href="http://dx.doi.org/10.5802/aif.1348">Gaps between consecutive divisors of factorials</a>, Ann. Inst. Fourier, 43 (3) (1993), 569-583.
%H Paul Erdős, S. W. Graham, Alexsandr Ivić, and Carl Pomerance, <a href="http://people.cst.cmich.edu/graha1sw/Pub/Papers/divfactorial.pdf">On the number of divisors of n!</a>, 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.
%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.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Complete_sequence">Complete sequence</a>.
%H <a href="/index/Fa#factorial">Index entries for sequences related to factorial numbers</a>
%H <a href="/index/Di#divisors">Index entries for sequences related to divisors of numbers</a>
%F a(n) <= a(n+1) <= 2*a(n) - _Benoit Cloitre_, Sep 07 2002
%F From Avik Roy (avik_3.1416(AT)yahoo.co.in), Jan 28 2009: (Start)
%F Assume, p1,p2...pm are the prime numbers less than or equal to n.
%F Then, a(n) = Product_{i=1..m} (bi+1), where bk = Sum_{i=1..m} floor(n/pk^i).
%F For example, if n=5, p1=2,p2=3,p3=5;
%F b1=floor(5/2)+floor(5/2^2)+floor(5/2^3)+...=2+1+0+..=3 similarly, b2=b3=1;
%F Thus a(5)=(3+1)(1+1)(1+1)=16. (End)
%F a(n) = A000005(A000142(n)). - _Michel Marcus_, Sep 13 2014
%F a(n) ~ exp(c * n/log(n) + O(n/log(n)^2)), where c = A131688 (Erdős et al., 1996). - _Amiram Eldar_, Nov 07 2020
%e a(4) = 8 because 4!=24 has precisely eight distinct divisors: 1, 2, 3, 4, 6, 8, 12, 24.
%p A027423 := n -> numtheory[tau](n!);
%t Table[ DivisorSigma[0, n! ], {n, 0, 35}]
%o (PARI) for(k=0,50,print1(numdiv(k!),", ")) \\ _Jaume Oliver Lafont_, Mar 09 2009
%o (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
%o (Haskell)
%o a027423 n = f 1 $ map (\p -> iterate (* p) p) a000040_list where
%o f y ((pps@(p:_)):ppss)
%o | p <= n = f (y * (sum (map (div n) $ takeWhile (<= n) pps) + 1)) ppss
%o | otherwise = y
%o -- _Reinhard Zumkeller_, Feb 27 2013
%o (Python 3.8+)
%o from math import prod
%o from collections import Counter
%o from sympy import factorint
%o 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
%Y Cf. A000005, A000142, A062569, A131688, A161466 (divisors of 10!).
%K nonn,easy,nice
%O 0,3
%A Glen Burch (gburch(AT)erols.com), _Leroy Quet_.
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