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 A085829 a(n) = least k such that the average number of divisors of {1..k} is >= n. 5
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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 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) EXAMPLE a(20) = 415739030 because the average number of divisors of {1..415739030} is >= 20. MATHEMATICA s = 0; k = 1; Do[ While[s = s + DivisorSigma[0, k]; s < k*n, k++ ]; Print[k]; k++, {n, 1, 20}] PROG (PARI) A085829(n) = {local(s, k); s=1; k=1; while(s

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Last modified July 20 01:38 EDT 2019. Contains 325168 sequences. (Running on oeis4.)