

A092669


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


11



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
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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 timeconsuming 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), 780782.
Index entries for sequences related to Egyptian fractions


FORMULA

a(n) = A092670(n)  A092670(n1).


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=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 *)


CROSSREFS

Cf. A092666, A092667, A092670, A092671, A092672, A006585.
Sequence in context: A127775 A210953 A254280 * A255986 A011400 A115013
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



