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Least k such that d(k) > product_{i=1..n} d(k-i)*d(k+i), where d(k) is the number of divisors of k.
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%I #9 Jul 21 2015 11:31:49

%S 12,107100,5055850800

%N Least k such that d(k) > product_{i=1..n} d(k-i)*d(k+i), where d(k) is the number of divisors of k.

%C Erdos and Suranyi prove that for every n there is a k such that d(k) > product_{i=1..n} d(k-i)*d(k+i). This sequence shows the least k for each n.

%C a(4) <= 205841401769203200. The number of divisors of a(4) must be > 36864. a(5) > 15824908106788068288000. The number of divisors of a(5) must be > 589824. [From _Donovan Johnson_, Mar 18 2010]

%D Paul Erdős and Janos Suranyi, Topics in the Theory of Numbers, Springer, 2003, p. 247.

%e a(2) = 107100 because 107100 is the least number that satisfies the inequality for n = 2. Note d(107100) = 108, d(107101) = d(107099) = 2 and d(107102) = d(107098) = 4.

%t k = 1; Table[dc=Table[DivisorSigma[0, k+i], {i, 0, 2n}]; k=k+n; While[dc[[n+1]]^2 <= Times@@dc, k++; dc=RotateLeft[dc]; dc[[2n+1]]=DivisorSigma[0, k+n]]; Print[{k, dc}]; k, {n, 2}]

%Y Cf. A000005 (number of divisors of n).

%K more,nonn,bref

%O 1,1

%A _T. D. Noe_, Feb 08 2005