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 A113648 A variant of Josephus Problem in which 2 persons are to be eliminated at the same time. 4
 1, 3, 6, 1, 3, 5, 7, 9, 12, 15, 18, 21, 24, 27, 30, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 1, 3, 5, 7 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS a(n) is defined as follows. Write the numbers 1 through 2n in a circle, start at 1 and n+1. Cross off every other number until only one number is left. The process that starts with 1 should be the first at any stage. For example we cross off 2, n+2, 4, n+4, 6, n+6, .... The remaining number is a(n). This function is defined only for even arguments. REFERENCES R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley LINKS FORMULA The sequence a(m) is defined for any even number m as follows: a(2) = 1. a(4*n) = 2*a(2*n) - 2*n - 1 (if a(2*n) > n) and a(4*n) = 2*a(2*n) + 2*n - 1 (if a(2*n) <= n). a(4*n+2) = 2*a(2*n+2) - 2*n - 5 (if a(2*n+2) >= n + 3), a(4*n+2) = 2*a(2*n+2) + 2*n - 2 (if n + 3 > a(2*n+2) >= 2), and a(4*n+2) = 2*n+1 (if a(2*n+2) = 1). EXAMPLE For a(8): we are to cross off 2, 6, 4, 8, 7, 3, 5 and 1 is left. Therefore a(8) = 1. MATHEMATICA jose2 = 1; jose2[n_] := If[Mod[n, 4] == 0, If[jose2[n/2] <= (n/4), 2(n/4) + 2jose2[n/2] - 1, 2jose2[n/2] - 2(n/4) - 1], Which[jose2[(n + 2)/2] == 1, n/2, 1 < jose2[(n + 2)/2] < (n + 10)/4, 2jose2[(n + 2)/2] + (n - 2)/2 - 2, (n + 6)/4 < jose2[(n + 2)/2], 2jose2[(n + 2)/2] - (n + 8)/2]]; CROSSREFS Cf. A006257. Sequence in context: A102257 A091425 A205005 * A104612 A088392 A195699 Adjacent sequences:  A113645 A113646 A113647 * A113649 A113650 A113651 KEYWORD easy,nonn AUTHOR Satoshi Hashiba, Daisuke Minematsu and Ryohei Miyadera, Jan 15 2006 STATUS approved

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Last modified April 17 08:16 EDT 2021. Contains 343064 sequences. (Running on oeis4.)