A103482 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.

12, 107100, 5055850800

Offset

1,1

Comments

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.

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]

References

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

Links

Table of n, a(n) for n=1..3.

Example

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.

Mathematica

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}]

Crossrefs

Cf. A000005 (number of divisors of n).

Sequence in context: A289295 A289569 A051368 * A056540 A331093 A369523

Adjacent sequences: A103479 A103480 A103481 * A103483 A103484 A103485

Keyword

more,nonn,bref

Author

T. D. Noe, Feb 08 2005

Status

approved