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Search: id:A125601
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| A125601 |
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a(n) = smallest k such that there are exactly n numbers whose sum of proper divisors is k. |
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+0
4
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| 2, 3, 6, 21, 37, 31, 49, 79, 73, 91, 115, 127, 151, 121, 181, 169, 217, 265, 253, 271, 211, 301, 433, 379, 331, 361, 457, 391, 451, 655, 463, 541, 421, 775, 511, 769, 673, 715, 865, 691, 1015, 631, 1069, 1075, 721, 931, 781, 1123, 871, 925, 901, 1177, 991, 1297
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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Minimal values for nodes of exact degree in aliquot sequences. Find each node's degree (number of predecessors) in aliquot sequences, and choose the smallest value as the sequence member. - Ophir Spector, ospectoro (AT) yahoo.com Nov 25 2007
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LINKS
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Ophir Spector, Table of n, a(n) for n = 0..157
W. Creyaufmueller, Aliquot sequences
MathWorld, Aliquot sequence
J. M. Pedersen, Tables of Aliquot Cycles
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EXAMPLE
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a(4) = 37 since there are exactly four numbers (155, 203, 299, 323) whose sum of proper divisors is 37. For k < 37 there are either fewer ore more numbers (32, 125, 161, 209, 221 for k = 31) whose sum of proper divisors is k.
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PROGRAM
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(PARI) {m=54; z=1500; y=600000; v=vector(z); for(n=2, y, s=sigma(n)-n; if(s<z, v[s]++)); w=vector(m, i, -1); for(j=2, z, if(v[j]<m&&w[v[j]+1]<0, w[v[j]+1]=j)); for(j=1, m, print1(w[j], ", "))}
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CROSSREFS
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Cf. A001065, A048138, A070015, A123930, A080907, A115350, A121507, A037020, A126016, A057709, A057710, A063990, A070015.
Sequence in context: A002078 A000372 A123930 this_sequence A025239 A127294 A012924
Adjacent sequences: A125598 A125599 A125600 this_sequence A125602 A125603 A125604
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KEYWORD
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nonn
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AUTHOR
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Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Nov 27 2006
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