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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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`%I #7 Mar 30 2012 17:22:32
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`%S 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,
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`%T 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,
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`%U 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
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`%N 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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`%C 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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`%H Peter Pein, <a href="/A093407/b093407.txt">Table of n, a(n) for n = 1..10515</a>
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`%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/EgyptianFraction.html">Egyptian Fraction</a>
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`%e a(1) = 3 because 2 = prime(1) and 1/1 + 1/3 = 4/3, whose numerator is divisible by 2.
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`%t 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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`%Y Cf. A092671 (n such that there is an Egyptian fraction partition of unity having smallest unit fraction 1/n), A093408.
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`%K nonn
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`%O 1,1
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`%A _T. D. Noe_, Mar 29 2004
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