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 A215721 The values of N for which the 1 X N domino-covering game is a second player win. 1

%I

%S 0,1,5,9,15,21,25,29,35,39,43,55,59,63,73,77,89,93,97,107,111,123,127,

%T 131,141,145,157,161,165,175,179,191,195,199,209,213,225,229,233

%N The values of N for which the 1 X N domino-covering game is a second player win.

%C In a domino-covering game, the players take turns placing dominoes (1 X 2 rectangles) on the board (here a 1 X N rectangle), such that the position of the dominoes is integer, and no two dominoes overlap. The loser is the first player unable to move.

%C The optimal strategies for this game can be determined by computing the nim-value nv(N) of each board 1 X N:

%C nv(0) = 0

%C nv(1) = 0

%C nv(N+2) = least nonnegative integer not in {nimsum(nv(k),nv(N-k)) : k <= N}

%C (where nimsum(a,b) is the bitwise xor of a and b).

%C The second player wins a game iff its nim-value is 0.

%H Charles R Greathouse IV, <a href="/A215721/b215721.txt">Table of n, a(n) for n = 1..10000</a>

%H Pratik Alladi, Neel Bhalla, Tanya Khovanova, Nathan Sheffield, Eddie Song, William Sun, Andrew The, Alan Wang, Naor Wiesel, Kevin Zhang Kevin Zhao, <a href="https://arxiv.org/abs/1707.07201">PRIMES STEP Plays Games</a>, arXiv:1707.07201 [math.CO], 2017, Section 8.

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,1,-1).

%F For n > 14, a(n) = a(n-5) + 34. - _Rainer Rosenthal_ and _Charles R Greathouse IV_, Sep 20 2012

%F G.f.: x^2*(6*x^13 - 2*x^12 + 8*x^10 - 2*x^9 - 2*x^8 + 2*x^7 + 3*x^5 + 6*x^4 + 6*x^3 + 4*x^2 + 4*x + 1) / ((x - 1)^2*(x^4 + x^3 + x^2 + x + 1)). - _Colin Barker_, Jan 26 2013

%e N=1: the first player is unable to move (second player win).

%e N=2,3: any move by the first player renders the second player unable to move (first player win).

%e N=4: the first player can win by covering the two central squares (first player win).

%e N=5: any move by the first player has a final counter-move by the second (second player win).

%o (Python)

%o import numpy as np

%o N = np.array([0,0])

%o U = np.arange(1000)

%o for i in U:

%o ..N = np.append(N, np.setdiff1d(U,np.bitwise_xor(N[:-1],N[-2::-1])).min())

%o print list(*np.where(N==0))

%o (PARI) a(n)=if(n<10,[0,1,5,9,15,21,25,29,35][n],n\5*34-[29,25,13,9,5][n%5+1]) \\ _Charles R Greathouse IV_, Aug 24 2012

%o (PARI) list(lim)=my(v=vector(lim\1+1),u);for(n=0,#v-3,u=vecsort(vector(n\2+1,k,bitxor(v[k],v[n-k+2])),,8);for(i=0,#u-1,if(u[i+1]!=i,v[n+3]=i;next(2));v[n+3]=#u));for(i=0,#v-1,v[i+1]=if(v[i+1],1,i)); vecsort(v,,8) \\ _Charles R Greathouse IV_, Aug 24 2012

%Y The nim-values are A002187(n-1).

%K nonn,easy

%O 1,3

%A _Yuval Gabay_, Aug 22 2012

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Last modified March 29 14:58 EDT 2020. Contains 333107 sequences. (Running on oeis4.)