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Search: id:A092669
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| A092669 |
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a(n) = number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k (for any k), 0<x_1<...<x_k=n. |
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+0
6
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| 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; listen)
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OFFSET
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1,15
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COMMENT
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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 (noe(AT)sspectra.com), Mar 26 2004
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REFERENCES
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Harry Ruderman and Paul Erdos, Problem E2427: Bounds of Egyptian fraction partitions of unity, Amer. Math. Monthly, Vol. 81, No. 7 (1974), 780-782.
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LINKS
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Toshitaka Suzuki, Table of n, a(n) for n = 1..610
Index entries for sequences related to Egyptian fractions
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FORMULA
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a(n)=A092670(n)-A092670(n-1)
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EXAMPLE
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a(6)=1 since there is the only fraction 1=1/2+1/3+1/6.
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MATHEMATICA
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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 (from T. D. Noe)
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CROSSREFS
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Cf. A092666, A092667, A092670, A092671, A092672, A006585.
Sequence in context: A108579 A143044 A127775 this_sequence A011400 A115013 A072736
Adjacent sequences: A092666 A092667 A092668 this_sequence A092670 A092671 A092672
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KEYWORD
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nonn
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AUTHOR
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Max Alekseyev (maxale(AT)gmail.com), Mar 02 2004
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EXTENSIONS
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More terms from T. D. Noe (noe(AT)sspectra.com), Mar 26 2004
More terms from T. Suzuki (suzuki(AT)scio.co.jp), Nov 24 2006
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