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Search: id:A061243
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| A061243 |
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a(n) = n+r where r is the smallest number such that n divides (n+1)(n+2)(n+3)...(n+r). |
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+0
4
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| 2, 4, 6, 8, 10, 9, 14, 12, 15, 15, 22, 16, 26, 21, 20, 22, 34, 24, 38, 25, 28, 33, 46, 28, 35, 39, 36, 35, 58, 35, 62, 40, 44, 51, 42, 42, 74, 57, 52, 45, 82, 49, 86, 55, 51, 69, 94, 54, 63, 60, 68, 65, 106, 63, 66, 63, 76, 87, 118, 65, 122, 93, 70, 72, 78, 77, 134, 85, 92
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Here r in most of the cases comes out to be Smarandache number given by A002034. It is conjectured that a(n) = n + S(n)
Dean Hickerson (dean.hickerson(AT)yahoo.com) remarks that the above conjecture is easy to prove: (n+1)(n+2)...(n+r) is congruent to r! (mod n), so n divides (n+1)(n+2)...(n+r) if and only if n divides r!. Oct 07, 2001
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EXAMPLE
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a(6) = 9 as 6 divides 7 X 8 X 9 but does not divide 7 X 8.
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MATHEMATICA
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a[n_] := For[r=1, True, r++, If[Mod[r!, n]==0, Return[n+r]]]
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CROSSREFS
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a(n) = n + A002034(n)
Sequence in context: A077127 A082191 A055952 this_sequence A079431 A081472 A097660
Adjacent sequences: A061240 A061241 A061242 this_sequence A061244 A061245 A061246
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KEYWORD
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nonn
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AUTHOR
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Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Apr 23 2001
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