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

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

1,15

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.

Index entries for sequences related to Egyptian fractions

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<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 *)

Crossrefs

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

Sequence in context: A127775 A210953 A254280 * A255986 A011400 A362271

Adjacent sequences: A092666 A092667 A092668 * A092670 A092671 A092672

Keyword

nonn

Author

Max Alekseyev, Mar 02 2004

Extensions

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

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

Status

approved