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For p = prime(n), the least k such that p is the numerator of a sum 1/k + 1/x1 +...+ 1/xm, where x1,...,xm (for any m) are distinct positive integers < k.
1

%I #4 Mar 30 2012 17:22:32

%S 6,2,3,4,3,4,5,4,5,5,5,6,6,8,5,7,7,5,6,7,8,8,7,7,6,8,7,5,8,8,6,7,5,8,

%T 8,10,8,9,9,7,8,10,9,9,10,8,7,7,8,8,10,8,8,9,10,9,8,8,9,9,8,7,9,8,10,

%U 7,10,9,10,10,8,9,8,10,9,10,7,9,9,11,10,9,9,10,10,9,10,7,9,9,11,10,9,14,14

%N For p = prime(n), the least k such that p is the numerator of a sum 1/k + 1/x1 +...+ 1/xm, where x1,...,xm (for any m) are distinct positive integers < k.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/EgyptianFraction.html">Egyptian Fraction</a>

%e a(1) = 6 because 2 = prime(1) and 1/2 + 1/6 = 2/3, whose numerator is 2.

%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, p]>0, a[[i]]=n, done=False]], {i, len}]]; a

%Y Cf. A093407.

%K nonn

%O 1,1

%A _T. D. Noe_, Mar 29 2004