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 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,15 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 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. FORMULA a(n) = A092670(n) - A092670(n-1). EXAMPLE a(6) = 1 since there is the only fraction 1 = 1/2+1/3+1/6. 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[i

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Last modified December 1 23:44 EST 2022. Contains 358485 sequences. (Running on oeis4.)