A093407 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.

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

Offset

1,1

Comments

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.

Links

Peter Pein, Table of n, a(n) for n = 1..10515

Eric Weisstein's World of Mathematics, Egyptian Fraction

Example

a(1) = 3 because 2 = prime(1) and 1/1 + 1/3 = 4/3, whose numerator is divisible by 2.

Mathematica

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

Crossrefs

Cf. A092671 (n such that there is an Egyptian fraction partition of unity having smallest unit fraction 1/n), A093408.

Sequence in context: A028292 A256244 A233386 * A349351 A147658 A221529

Adjacent sequences: A093404 A093405 A093406 * A093408 A093409 A093410

Keyword

nonn

Author

T. D. Noe, Mar 29 2004

Status

approved