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%I #18 Jan 12 2016 06:10:43
%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 time-consuming 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 Erdős, <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), 780-782.
%H <a href="/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>
%F a(n) = A092670(n) - A092670(n-1).
%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=n-1; 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
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