

A206602


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.


2



2, 5, 11, 14, 36, 57, 141, 221, 346, 677, 4042, 9870, 114916, 179557, 1070250, 2612917, 9967491, 12459364, 19467757, 30418371, 38022964, 59410882, 116036880, 283293166, 553306966, 864542135, 1080677669, 3297966522, 8051676081, 15725929847, 19657412309, 47991729272
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OFFSET

1,1


COMMENTS

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".
It gives sequences for every step length s from 2.
s=2: 2^n  1;
s=3: 3, 5, 8, 30, 69, 104, 354, 798, 1797, 2696, 9102, 20481.
s=5: this sequence.
Apparently, this is Seki Takakazu's sequence of "limitative numbers" with m = 5.  Petros Hadjicostas, Jul 18 2020


LINKS

Saburô Uchiyama, On the generalized Josephus problem, Tsukuba J. Math. 27(2) (2003), 319339; see p. 337. [Has about 50 sequences related to Seki Takakazu's "limitative numbers"]


EXAMPLE

The solution for a(3) = 11: (WBWBWBWBWWBBBWWWBBBWBW);
White stones: (5, 10, 15, 20, 3, 9, 16, 22, 7, 14, 1);
Black stones: (8, 18, 4, 17, 6, 21, 13, 12, 20, 2, 11).


MAPLE

s:=5: s1:=s1: a:=1:
for p from 2 to 100000 by 2 do
b:=(a+s1) mod p +1:
if (b=1) then printf("%9d", p1): end if:
a:=(b+s1) mod (p+1) +1:
if (a=1) then printf("%9d", p): end if:
end do:


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



