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Search: id:A117846
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| A117846 |
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Given n, define a(n) as follows: let a sequence b(k) be defined by b(k+1)=b(k)+b(k)mod k; b(1)=2n-1. (Here b(k)mod k denotes the least nonnegative residue of b(k) modulo k). Let a(n) be the common value of b(k+1)-b(k) for all large k if such exists; otherwise let a(n) be 0. |
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| 97, 1, 2, 2, 316, 2, 3, 3, 3, 4, 12, 4, 4, 12, 11, 11, 316, 11, 316, 316, 6, 316, 316, 316, 316, 97, 316, 316, 13, 316, 13, 13, 13, 13, 8, 13, 13, 12, 13, 13, 13, 13, 13, 13, 14, 14, 316, 14, 316, 316, 316, 97, 9, 97, 97, 13, 10, 10, 11, 10, 14, 11, 12, 12, 97, 12, 97, 132
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Putting b(1)=2n gives essentially the same sequence as putting b(1)=2n-1. It is a plausible conjecture or at least an interesting open problem that a(n) is never zero; that is all the sequences b(k) are arithmetic progressions from some point on. Sequence A073117 is the sequence b(k) with b(1)=1. Do the values a(n) include all positive numbers?
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EXAMPLE
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n=4: b(1)=7 and the sequence b(k) continues 7,8,10,12,14...with b(k+1)-b(k)=2 for all k>3, so a(4)=2.
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CROSSREFS
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Cf. A073117.
Sequence in context: A008702 A133402 A163494 this_sequence A058286 A051330 A106429
Adjacent sequences: A117843 A117844 A117845 this_sequence A117847 A117848 A117849
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KEYWORD
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nonn
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AUTHOR
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Alex Abercrombie (aabercrombie(AT)whsmithnet.co.uk), Mar 22 2007
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