A085829 a(n) = least k such that the average number of divisors of {1..k} is >= n.

1, 4, 15, 42, 120, 336, 930, 2548, 6930, 18870, 51300, 139440, 379080, 1030484, 2801202, 7614530, 20698132, 56264040, 152941824, 415739030, 1130096128, 3071920000, 8350344420, 22698590508, 61701166395, 167721158286, 455913379324, 1239301050624, 3368769533514

Offset: 1

Author

Robert G. Wilson v, Jul 07 2003

Keywords

• nonn
• nice

Comments

Does a(n+1)/a(n) converge to e?

Since the total number of divisors of {1..k} (see A006218) is k * (log(k) + 2*gamma - 1) + O(sqrt(k)), the average number of divisors of {1..k} approaches (log(k) + 2*gamma - 1). Since log(a(n)) + 2*gamma - 1 approaches n, a(n+1)/a(n) approaches e. - Jon E. Schoenfield, Aug 13 2007

Examples

			a(20) = 415739030 because the average number of divisors of {1..415739030} is >= 20.
		

References

• Julian Havil, "Gamma: Exploring Euler's Constant", Princeton University Press, Princeton and Oxford, pp. 112-113, 2003.

Links

• Donovan Johnson, Table of n, a(n) for n = 1..40 (first 36 terms from Jon E. Schoenfield)

Crossrefs

Cf. A050226, A057494, A085567, A006218.

Programs

Mathematica

s = 0; k = 1; Do[ While[s = s + DivisorSigma[0, k]; s < k*n, k++ ]; Print[k]; k++, {n, 1, 20}]

PARI

A085829(n) = {local(s,k);s=1;k=1;while(sMichael B. Porter, Oct 23 2009

Extensions

Edited by Don Reble, Nov 06 2005

More terms from Jon E. Schoenfield, Aug 13 2007