A137607 -id:A137607 - cp's OEIS frontend

A096259 Longest period of an abstract version of the game of Go on a 1 X n board.

1, 2, 6, 24, 70, 180, 294, 112, 270, 900, 330, 792

Offset: 1

Yasutoshi Kohmoto, Aug 01 2004; revised Apr 23 2008

Rules: 1. If a set of a player's stones has no "open edge" then the other player get the set of stones.
2. If the sets of both player's stones has no "open edge" in a configuration, then a player who made this configuration get the set of the other player's stone.
3. A player never make a configuration in which his stones have no open edge and the other player's stones have an open edge.
A board is represented as follows.
  + + + +
  + o x +
  + + + +
"o" means a white stone, "x" means a black stone.
"Open edge" : An edge which has one node without a stone. Example:
  + x x +
  x o o x
  + x x +
The center set of white stones has no "open edge", so black player gets them. Six black stones have "open edges" like this : "x +".
Note that the rules do not specify when a player wins, so the game never terminates.

			The case n=3:
  t 1 2 3 3 4 4 5 6 6 7 7
  + x x x x x + x x + x x
  + + + x x x + + o o o +
  + + o o + o o o o o o +
t=1 and t=7 are the same, so the period is 6.
a(12) = 12 * 2^0 * (12 + 6 + 3 + 10 + 5 + 9 + 7 + 8 + 4 + 2 + 1 - 1) = 792.

Cf. A007565, A048289, A137604, A137605, A137606, A137607.

For 4<=n, a(n) = n * 2^p * ( Sum_{0<=k<=m} ( Sum_{0<=i<=h_k} n_k/2^i ) - 1 ) where p = m Mod 2, n_0 = n, n_k = n - [n_{k-1}/2^(h_{k-1}+1)] - 1, 2^h_k is the highest power of two dividing n_k: n_m/2^h_m = 1.

A137606 Numbers m such that all numbers {1...m} appear in the sequence {b(0) = m, b(n+1) = b(n)/2 if even, m-(b(n)+1)/2 otherwise}.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 10, 12, 15, 19, 24, 27, 30, 31, 34, 36, 40, 42, 51, 52, 54, 66, 70, 75, 82, 84, 87, 90, 91, 96, 99, 100, 106, 114, 120, 132, 135, 136, 147, 156, 159, 174, 175, 180, 184, 187, 190, 192, 195, 210, 211, 222, 231, 232, 234, 240, 244, 246, 252, 255, 262

Offset: 1

Yasutoshi Kohmoto, Apr 23 2008

nonn

Lemma: A sequence {b(n)} defined as above with m>1 cannot have values outside [1,m]. (For m=1, b=(1,0,0,0....).)
Corollary: Such a sequence {b(n)} is periodic with period <= m (except maybe for some initial terms).
Lemma 2: For any m>1, b(1) = floor( m/2 ) and if b(n)=m-1, then b(n+1)= [ m/2 ].
Proposition: As soon as there is a term b(n)=2^k, the (b-)sequence continues b(n+1)=2^(k-1),...,b(n+k)=1, b(n+k+1)=m-1 and then starts over with b(n+k+2)=b(1).
Corollary 2: Numbers m=2^k, k>2 cannot appear in the present sequence.
Proposition: For any b(0)=m>1, sooner or later the value 1 is reached.
Generate a sequence b(n) by the following rule. If b(n-1) is divisible by 2 then b(n) = b(n-1)/2. If b(n-1) is not divisible by 2 then b(n) = b(0)-(b(n-1)+1)/2. When b(n)=1 it ends. Sequence gives all m such that all numbers k with 1<=k<=m-2 appear in b(n), b(0)=m.
Sequence contains 1 and numbers m>1 such that 2m-1 is prime and -2 or 2 is a primitive root modulo 2m-1. - Max Alekseyev, May 16 2008

			6->3->4->2->1. 1,2,3,4=6-2 appear in b(n), b(0)=6. So 6 is a term of A137606.

Robert G. Wilson v, Table of n, a(n) for n = 1..2354.

Cf. A137607, A096259, A105874, A001122.

f[n_] := Block[{lst = {n}, a}, While[a = Last@ lst; a != 1, AppendTo[lst, If[ EvenQ@ a, a/2, lst[[1]] - (a + 1)/2]]]; Length@ lst - 1]; t = Array[f, 262]; Select[ Range @ 262, t[[ # ]] == # - 2 &] (* Robert G. Wilson v *)

b137606(n)= n=[n]; for( i=1,n[1]-1, n=concat( n, if( n[i]%2, n[1]-(n[i]+1)/2, n[i]/2 )); n[i]>1 || break); n
A137606(Nmax) = for( n=1,Nmax, n==#b137606(n) && print1(n","))

forprime(p=3,10^3, if(znorder(Mod(-2,p))==p-1||znorder(Mod(2,p))==p-1, print1((p+1)/2,", ") )) \\ Max Alekseyev, May 16 2008

This sequence as a set is the union of { 1 }, { (A105874(n)+1)/2 } and { (A001122(n)+1)/2 }. - Max Alekseyev, May 16 2008

Edited & extended by M. F. Hasler, Apr 28 2008

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A096259 Longest period of an abstract version of the game of Go on a 1 X n board.

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A137606 Numbers m such that all numbers {1...m} appear in the sequence {b(0) = m, b(n+1) = b(n)/2 if even, m-(b(n)+1)/2 otherwise}.

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