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a(n) = Max_{k=1..n} tau(k) where tau(x)=A000005(x) is the number of divisors of x.
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%I #62 Mar 06 2023 02:58:36

%S 1,2,2,3,3,4,4,4,4,4,4,6,6,6,6,6,6,6,6,6,6,6,6,8,8,8,8,8,8,8,8,8,8,8,

%T 8,9,9,9,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,10,10,12,12,

%U 12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12

%N a(n) = Max_{k=1..n} tau(k) where tau(x)=A000005(x) is the number of divisors of x.

%C Is this the same as A068509? - _David Scambler_, Sep 10 2012

%C They are different even asymptotically: A068509(n)=O(sqrt(n)), while a(n) does not have polynomial growth. One example where the sequences differ: a(625) = 24 < A068509(625). (The inequality is implied by the set {1,2,..,25} where each pair of the elements has lcm <= 625.) - _Max Alekseyev_, Sep 11 2012

%C The two sequences first differ when n = 336, due to the set of 21 elements {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 21, 24, 30, 36, 42, 48} where each pair of elements has lcm <= 336, while no positive integer <= 336 has more than 20 divisors. Therefore A068509(336) = 21 and A070319(336) = 20. - _William Rex Marshall_, Sep 11 2012

%C Indices of records give A002182. - _Omar E. Pol_, Feb 18 2023

%D Sándor, J., Crstici, B., Mitrinović, Dragoslav S. Handbook of Number Theory I. Dordrecht: Kluwer Academic, 2006, p. 44.

%D S. Wigert, Sur l'ordre de grandeur du nombre des diviseurs d'un entier, Arkiv. for Math. 3 (1907), 1-9.

%H Charles R Greathouse IV, <a href="/A070319/b070319.txt">Table of n, a(n) for n = 1..10000</a>

%H S. Ramanujan, <a href="http://ramanujan.sirinudi.org/Volumes/published/ram15.html">Highly composite numbers</a>, Proceedings of the London Mathematical Society, 2, XIV, 1915, 347 - 409.

%F a(n) = exp(log(2) log(n) / log(log(n)) + O(log(n) log(log(log(n))) / (log(log(n)))^2)). (See Sándor reference for more formulas.) - _Eric M. Schmidt_, Jun 30 2013

%F a(n) = A002183(A261100(n)). - _Antti Karttunen_, Jun 06 2017

%t a = {0}; Do[AppendTo[a, Max[DivisorSigma[0, n], a[[n]]]], {n, 120}]; Rest@ a (* _Michael De Vlieger_, Sep 29 2015 *)

%o (PARI) a(n)=vecmax(vector(n,k,numdiv(k)))

%o (PARI) v=vector(100);v[1]=1;for(n=2,#v,v[n]=max(v[n-1],numdiv(n))); v \\ _Charles R Greathouse IV_, Sep 12 2012

%o (PARI) A070319(n,m=1,s=2)={for(k=s,n,m<numdiv(k) && m=numdiv(k));m} /* Although this should statistically require more assignments, the simple for() loop is faster than a forstep(k=n,s,-1) loop. To speed up the computation, give as 2nd and 3rd (optional) arguments earlier computed values, e.g. m=a(n-1) and s=n, cf. the example below. */ \\ _M. F. Hasler_, Sep 12 2012

%o (PARI) {a=0;for(n=1,100,print1(a=A070319(n,a,n),","))} /* Using this pattern, computation of a(1..10^6) is faster than "normal" computation of a(1..3000). */

%o (Haskell)

%o a070319 n = a070319_list !! (n-1)

%o a070319_list = scanl1 max $ map a000005 [1..]

%o -- _Reinhard Zumkeller_, Apr 01 2011

%Y Cf. A000005, A002182, A002183, A261100, A261104.

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, May 11 2002