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A028933 Table of winning positions in Tchoukaillon (or Mancala) solitaire.

%I A028933 #22 Jul 08 2025 18:57:00
%S A028933 0,1,0,2,1,2,0,1,3,1,1,3,0,0,2,4,1,0,2,4,0,2,2,4,1,2,2,4,0,1,1,3,5,1,
%T A028933 1,1,3,5,0,0,0,2,4,6,1,0,0,2,4,6,0,2,0,2,4,6,1,2,0,2,4,6,0,1,3,2,4,6,
%U A028933 1,1,3,2,4,6,0,0,2,1,3,5,7,1,0,2,1,3,5,7
%N A028933 Table of winning positions in Tchoukaillon (or Mancala) solitaire.
%C A028933 Table read by rows where b(n,i) = the number of counters in the i-th position from the store of the unique winning Tchoukaillon board having n total counters.
%H A028933 D. Betten, <a href="http://dx.doi.org/10.1016/S0167-5060(08)70224-3">Kalahari and the Sequence "Sloane No. 377"</a>, Annals Discrete Math., 37, 51-58, 1988.
%H A028933 D. M. Broline and _Daniel E. Loeb_, <a href="http://arXiv.org/abs/math.CO/9502225">The combinatorics of Mancala-Type games: Ayo, Tchoukaillon and 1/Pi</a>, J. Undergrad. Math. Applic., vol. 16 (1995), pp. 21-36.
%H A028933 Brant Jones, Laura Taalman and Anthony Tongen, <a href="http://www.jstor.org/stable/10.4169/amer.math.monthly.120.08.706">Solitaire Mancala Games and the Chinese Remainder Theorem</a>, Amer. Math. Monthly, 120 (2013), 706-724.
%F A028933 Let p(n) be the minimum j such that b(n,j) = 0.  (This is A028920.)
%F A028933 Directly from the rules of Tchoukaillon, we find b(n+1,i) = (b(n,i) - 1 for 1 <= i < p(n), i for i = p(n), and b(n,i) for i > p(n)).
%F A028933 Also, b(n,i) = (n - Sum_{j=1..(i-1)} b(n,j)) mod (i+1).
%e A028933    The rows of b(n,i) begin
%e A028933    n\i 1 2 3 4 5 6
%e A028933    1   1
%e A028933    2   0 2
%e A028933    3   1 2
%e A028933    4   0 1 3
%e A028933    5   1 1 3
%e A028933    6   0 0 2 4
%e A028933    7   1 0 2 4
%e A028933    8   0 2 2 4
%e A028933    9   1 2 2 4
%e A028933    10  0 1 1 3 5
%e A028933    11  1 1 1 3 5
%e A028933    12  0 0 0 2 4 6
%e A028933    13  1 0 0 2 4 6
%e A028933    14  0 2 0 2 4 6
%e A028933    15  1 2 0 2 4 6
%e A028933    16  0 1 3 2 4 6
%e A028933    17  1 1 3 2 4 6
%t A028933 s[list_] := Module[{x = Append[list, 0], i = 1}, While[x[[i]] =!= 0, x[[i]] = x[[i]] - 1; i = i + 1]; x[[i]] = i; If[Last@x == 0, Most[x], x]]; Prepend[Flatten@NestList[s, {}, 20],0] (* _Birkas Gyorgy_, Feb 26 2011 *)
%Y A028933 Cf. A002491, A028920, A007952, A028931, A028932, A028933.
%K A028933 nonn,tabl
%O A028933 0,4
%A A028933 _N. J. A. Sloane_
%E A028933 Formulas added by _Brant Jones_, Oct 14 2013