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%I #14 Jun 23 2020 11:45:09
%S 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,
%T 25,27,29,31,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,
%U 90,93,96,99,102,105,108,111,114,117,120,123,126,1,3,5,7
%N A variant of Josephus Problem in which 2 persons are to be eliminated at the same time.
%C 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.
%D R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley
%H <a href="/index/J#Josephus">Index entries for sequences related to the Josephus Problem</a>
%F 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).
%e For a(8): we are to cross off 2, 6, 4, 8, 7, 3, 5 and 1 is left. Therefore a(8) = 1.
%t jose2[2] = 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]];
%Y Cf. A006257.
%K easy,nonn
%O 1,2
%A Satoshi Hashiba, Daisuke Minematsu and _Ryohei Miyadera_, Jan 15 2006
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