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%I #38 Jan 19 2023 09:35:21
%S 2,5,11,14,36,57,141,221,346,677,4042,9870,114916,179557,1070250,
%T 2612917,9967491,12459364,19467757,30418371,38022964,59410882,
%U 116036880,283293166,553306966,864542135,1080677669,3297966522,8051676081,15725929847,19657412309,47991729272
%N a(n) is the number of white and black stones lying in a circle; starting with place 1, the first a(n) steps of length 5 give the places of white stones. Beginning with last place, the next a(n) steps give the places of black stones.
%C The game eliminates a(n) - 1 white and a(n) black stones; the a(n)-th white stone survives. The game is known under different names; e.g., "Sankt Peters Spiel" or "Ludus Sancti Petri" or "Josephus problem".
%C It gives sequences for every step length s from 2.
%C s=2: 2^n - 1;
%C s=3: 3, 5, 8, 30, 69, 104, 354, 798, 1797, 2696, 9102, 20481.
%C s=5: this sequence.
%C Apparently, this is Seki Takakazu's sequence of "limitative numbers" with m = 5. - _Petros Hadjicostas_, Jul 18 2020
%H R. Baumann, <a href="https://www.yumpu.com/de/document/read/460296/nr-165">Computer Knobelei</a>, LOG IN, Heft Nr. 165, pp. 68-71, 2010 (in German).
%H Saburô Uchiyama, <a href="https://doi.org/10.21099/tkbjm/1496164652">On the generalized Josephus problem</a>, Tsukuba J. Math. 27(2) (2003), 319-339; see p. 337. [Has about 50 sequences related to Seki Takakazu's "limitative numbers"]
%H Saburô Uchiyama, <a href="https://www.jstor.org/stable/43686318">On the generalized Josephus problem</a>, Tsukuba J. Math. 27(2) (2003), 319-339 [jstor stable version]; see p. 337.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Seki_Takakazu">Seki Takakazu</a>.
%H <a href="/index/J#Josephus">Index entries for sequences related to the Josephus Problem</a>
%e The solution for a(3) = 11: (WBWBWBWBWWBBBWWWBBBWBW);
%e White stones: (5, 10, 15, 20, 3, 9, 16, 22, 7, 14, 1);
%e Black stones: (8, 18, 4, 17, 6, 21, 13, 12, 20, 2, 11).
%p s:=5: s1:=s-1: a:=1:
%p for p from 2 to 100000 by 2 do
%p b:=(a+s1) mod p +1:
%p if (b=1) then printf("%9d",p-1): end if:
%p a:=(b+s1) mod (p+1) +1:
%p if (a=1) then printf("%9d",p): end if:
%p end do:
%K nonn
%O 1,1
%A _Paul Weisenhorn_, Feb 10 2012
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