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A007495
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Josephus problem: survivors.
(Formerly M0237)
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10
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1, 1, 2, 2, 2, 4, 5, 4, 8, 8, 7, 11, 8, 13, 4, 11, 12, 8, 12, 2, 13, 7, 22, 2, 8, 13, 26, 4, 26, 29, 17, 27, 26, 7, 33, 20, 16, 22, 29, 4, 13, 22, 25, 14, 22, 37, 18, 46, 42, 46, 9, 41, 12, 7, 26, 42, 24, 5, 44, 53, 52, 58, 29, 22, 12, 48, 27, 30, 58, 52, 49, 57, 13, 14, 32, 24, 75, 8, 67
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OFFSET
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1,3
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COMMENTS
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If, in a circle of k persons, every n-th person is removed, the survivor is t(k,n) + 1. So the recurrence generates a sequence of survivors. See the formula. For more details see the "Proof of the formula". - Gerhard Kirchner, Oct 23 2016
The recurrence formula looks like a simple congruential generator for pseudo-random numbers. Is a(n) pseudo-random? It seems so, see: "Stochastic aspects". I used the formula for extending a(n) up to n=2^20. - Gerhard Kirchner, Nov 10 2016
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REFERENCES
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Friend H. Kierstead, Jr., Computer Challenge Corner, J. Rec. Math., 10 (1977), see p. 124.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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Let t(k,n) = (t(k-1,n) + n) mod k and t(1,n) = 0; then a(n) = t(n,n) + 1. - Gerhard Kirchner, Oct 23 2016
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EXAMPLE
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If n = 4 we have that:
t(1,4) = 0.
t(2,4) = (0+4) mod 2 = 0.
t(3,4) = (0+4) mod 3 = 1.
t(4,4) = (1+4) mod 4 = 1.
So a(4) = 1 + 1 = 2. (End)
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MATHEMATICA
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(* First do *) Needs["Combinatorica`"] (* then *) f[n_] := Last@ InversePermutation@ Josephus[n, n]; Array[f, 80] (* Robert G. Wilson v, Jul 31 2010 *)
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CROSSREFS
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KEYWORD
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easy,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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