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Search: id:A085829
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| A085829 |
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a(n) = least k such that the average number of divisors of {1..k} is >= n. |
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+0
5
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| 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
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Does a(n+1)/a(n) converge to e?
Does a(n+1)/a(n) converge to e? Reply from Jon E. Schoenfield: 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 (jonscho(AT)hiwaay.net), Aug 13 2007
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REFERENCES
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Julian Havil, "Gamma: Exploring Euler's Constant", Princeton University Press, Princeton and Oxford, pp. 112-113, 2003.
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LINKS
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Jon Schoenfield, Table of n, a(n) for n=1..36
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EXAMPLE
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a(20) = 415739030 because the average number of divisors of {1..415739030} is >= 20.
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MATHEMATICA
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s = 0; k = 1; Do[ While[s = s + DivisorSigma[0, k]; s < k*n, k++ ]; Print[k]; k++, {n, 1, 20}]
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CROSSREFS
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Cf. A050226, A057494, A085567.
Cf. A006218.
Adjacent sequences: A085826 A085827 A085828 this_sequence A085830 A085831 A085832
Sequence in context: A011844 A075468 A100503 this_sequence A085567 A075673 A062827
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KEYWORD
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nonn,nice
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AUTHOR
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Robert G. Wilson v (rgwv(AT)rgwv.com), Jul 07 2003
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EXTENSIONS
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Edited by Don Reble (djr(AT)nk.ca), Nov 06 2005
More terms from Jon E. Schoenfield (jonscho(AT)hiwaay.net), Aug 13 2007
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