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A123859
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a(1)=1. a(n) = a(n-1) + (sum of the terms, from among terms a(1) through a(n-1), which divide sum{k=1 to n-1} a(k)).
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1
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1, 2, 3, 9, 13, 16, 19, 32, 52, 56, 57, 125, 126, 127, 130, 184, 243, 244, 245, 248, 254, 257, 258, 259, 278, 281, 294, 298, 299, 979, 980, 984, 1073, 1076, 1079, 1083, 1357, 1358, 1362, 1375, 1378, 1381, 1510, 1524, 1525, 1547, 1548, 1549, 1552, 1558, 1626
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OFFSET
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1,2
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LINKS
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EXAMPLE
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The sum of the first 7 terms of the sequence is 63. The terms, from among the first 7 terms, which divide 63 are 1,3,9. So a(8) = a(7) + 1 + 3 + 9 = 32.
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MATHEMATICA
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a = {1}; For[n = 2, n < 60, n++, su = Plus @@ a; co = 0; For[i = 1, i < n, i++, If[IntegerQ[su/a[[i]]], co = co + a[[i]]]]; AppendTo[a, a[[ -1]] + co]]; a (* Stefan Steinerberger, Jun 27 2007 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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