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Search: id:A093407
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| A093407 |
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For p = prime(n), the least k such that p divides the numerator of a sum 1/k + 1/x1 +...+ 1/xm, where x1,...,xm (for any m) are distinct positive integers < k. |
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+0
3
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| 3, 2, 3, 4, 3, 4, 5, 4, 5, 5, 5, 6, 6, 7, 5, 7, 7, 5, 6, 7, 7, 8, 7, 7, 6, 8, 7, 5, 8, 8, 6, 7, 5, 8, 8, 9, 8, 8, 9, 7, 8, 9, 9, 9, 9, 8, 7, 7, 8, 8, 10, 8, 8, 9, 10, 9, 8, 8, 9, 9, 8, 7, 9, 8, 10, 7, 9, 9, 10, 10, 8, 9, 8, 10, 9, 10, 7, 9, 9, 11, 9, 9, 9, 10, 10, 9, 10, 7, 9, 9, 11, 10, 9, 11, 11, 11
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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This is a very slow-growing sequence: for n <= 1000, a(n) <= 18. The number a(n) * prime(n) is the least number divisible by prime(n) in sequence A092671.
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LINKS
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Peter Pein, Table of n, a(n) for n = 1..10515
Eric Weisstein's World of Mathematics, Egyptian Fraction
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EXAMPLE
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a(1) = 3 because 2 = prime(1) and 1/1 + 1/3 = 4/3, whose numerator is divisible by 2.
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MATHEMATICA
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len=100; a=Table[0, {len}]; done=False; s={0}; n=0; While[ !done, n++; s=Join[s, s+1/n]; ns=Numerator[s]; done=True; Do[If[a[[i]]==0, p=Prime[i]; If[Count[ns, _?(#>0 && Mod[ #, p]==0&)]>0, a[[i]]=n, done=False]], {i, len}]]; a
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CROSSREFS
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Cf. A092671 (n such that there is an Egyptian fraction partition of unity having smallest unit fraction 1/n), A093408.
Sequence in context: A007888 A077178 A028292 this_sequence A105161 A094365 A098822
Adjacent sequences: A093404 A093405 A093406 this_sequence A093408 A093409 A093410
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KEYWORD
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nonn
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AUTHOR
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T. D. Noe (noe(AT)sspectra.com), Mar 29 2004
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