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Search: id:A059955
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| A059955 |
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a(n)= [p(n)!/lcm{1,2,...,p(n)}] modulo[p(n)], where p(n) = n-th prime. |
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2
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| 1, 2, 5, 10, 3, 10, 4, 3, 28, 17, 18, 30, 20, 41, 42, 14, 19, 30, 37, 63, 50, 7, 12, 83, 30, 91, 19, 69, 91, 97, 56, 22, 80, 39, 137, 44, 9, 154, 19, 37, 141, 141, 168, 126, 183, 200, 205, 136, 55, 95, 204, 126, 213, 230, 68, 63, 158, 202, 162, 102, 182, 104, 38, 165
(list; graph; listen)
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OFFSET
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2,2
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COMMENT
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a(n) gives also the smallest coefficient for which the multiple M of lcm(1 through p(n)-1) satisfies p(n) divides M + 1. This computes the solution of the puzzle requiring the smallest number such that grouping in 2's, 3's, etc. up to the n-th prime,all leave a remainder of one except the last which leaves no remainder.
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EXAMPLE
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a(7)=10 because p(7)=17 and [17!/lcm(1,2,...,17)] mod 17 = 29030400 mod 17 = 10
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MAPLE
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for n from 2 to 150 do printf(`%d, `, floor(ithprime(n)!/ilcm(i $ i=1..ithprime(n))) mod ithprime(n) ); od:
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CROSSREFS
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Cf. A025527, A003418.
Sequence in context: A109469 A083460 A125974 this_sequence A099796 A022831 A064365
Adjacent sequences: A059952 A059953 A059954 this_sequence A059956 A059957 A059958
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KEYWORD
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nonn
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AUTHOR
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Lekraj Beedassy (blekraj(AT)yahoo.com), Mar 13 2001
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EXTENSIONS
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More terms from James A. Sellers (sellersj(AT)math.psu.edu), Mar 15 2001
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