A092669 a(n) = number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k (for any k), 0

1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 3, 0, 0, 5, 0, 11, 0, 0, 0, 19, 0, 0, 0, 73, 0, 86, 0, 0, 163, 0, 203, 286, 0, 0, 0, 803, 0, 1399, 0, 0, 2723, 0, 0, 4870, 0, 0, 0, 8789, 0, 13937, 14987, 42081, 0, 0, 0, 85577, 0, 0, 159982, 0, 117889, 437874, 0, 0, 0, 818640, 0

Offset: 1

Author

Max Alekseyev, Mar 02 2004

Keywords

• nonn

Comments

For a given n, the Mathematica program uses backtracking to count the solutions. The solutions can be printed by uncommenting the print statement. It is very time-consuming for large n. A092671 gives the n that yield a(n) > 0. - T. D. Noe, Mar 26 2004

Examples

			a(6) = 1 since there is the only fraction 1 = 1/2+1/3+1/6.
		

Links

• Toshitaka Suzuki, Table of n, a(n) for n = 1..610
• Harry Ruderman and Paul Erdős, Problem E2427: Bounds of Egyptian fraction partitions of unity, Amer. Math. Monthly, Vol. 81, No. 7 (1974), 780-782.
• Index entries for sequences related to Egyptian fractions

Crossrefs

Cf. A092666, A092667, A092670, A092671, A092672, A006585.

Programs

Mathematica

n=20; try2[lev_, s_] := Module[{nmim, nmax, si, i}, AppendTo[soln, 0]; If[lev==1, nmin=2, nmin=1+soln[[ -2]]]; nmax=n-1; Do[If[iT. D. Noe, Mar 26 2004 *)

Formula

a(n) = A092670(n) - A092670(n-1).

Extensions

More terms from T. D. Noe, Mar 26 2004

More terms from T. Suzuki (suzuki(AT)scio.co.jp), Nov 24 2006