%I
%S 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,
%T 163,0,203,286,0,0,0,803,0,1399,0,0,2723,0,0,4870,0,0,0,8789,0,13937,
%U 14987,42081,0,0,0,85577,0,0,159982,0,117889,437874,0,0,0,818640,0
%N a(n) = number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k (for any k), 0<x_1<...<x_k=n.
%C 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 timeconsuming for large n. A092671 gives the n that yield a(n) > 0.  _T. D. Noe_, Mar 26 2004
%H Toshitaka Suzuki, <a href="/A092669/b092669.txt">Table of n, a(n) for n = 1..610</a>
%H Harry Ruderman and Paul ErdÅ‘s, <a href="http://www.jstor.org/stable/2319578">Problem E2427: Bounds of Egyptian fraction partitions of unity</a>, Amer. Math. Monthly, Vol. 81, No. 7 (1974), 780782.
%H <a href="/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>
%F a(n) = A092670(n)  A092670(n1).
%e a(6) = 1 since there is the only fraction 1 = 1/2+1/3+1/6.
%t n=20; try2[lev_, s_] := Module[{nmim, nmax, si, i}, AppendTo[soln, 0]; If[lev==1, nmin=2, nmin=1+soln[[ 2]]]; nmax=n1; Do[If[i<n/2  !PrimeQ[i], si=s+1/i; If[si==1, soln[[ 1]]=i; (*Print[soln];*) cnt++ ]; If[si<1, soln[[ 1]]=i; try2[lev+1, si]]], {i, nmin, nmax}]; soln=Drop[soln, 1]]; soln={n}; cnt=0; try2[1, 1/n]; cnt (* _T. D. Noe_, Mar 26 2004 *)
%Y Cf. A092666, A092667, A092670, A092671, A092672, A006585.
%K nonn
%O 1,15
%A _Max Alekseyev_, Mar 02 2004
%E More terms from _T. D. Noe_, Mar 26 2004
%E More terms from T. Suzuki (suzuki(AT)scio.co.jp), Nov 24 2006
