

A159257


Rank deficiency of the Lights Out problem of size n.


9



0, 0, 0, 4, 2, 0, 0, 0, 8, 0, 6, 0, 0, 4, 0, 8, 2, 0, 16, 0, 0, 0, 14, 4, 0, 0, 0, 0, 10, 20, 0, 20, 16, 4, 6, 0, 0, 0, 32, 0, 2, 0, 0, 4, 0, 0, 30, 0, 8, 8, 0, 0, 2, 4, 0, 0, 0, 0, 22, 0, 40, 24, 0, 28, 42, 0, 32, 0, 8, 0, 14, 0, 0, 4, 0, 0, 2, 0, 64, 0, 0, 0, 6, 12, 0, 0, 0, 0, 10, 0, 0, 20, 0, 4, 62, 0, 0, 20, 16, 0, 18, 0, 0, 4, 0, 0, 6, 0, 8, 0, 0, 0, 2, 4, 0, 0, 0, 8, 46, 0, 0, 0, 80, 4, 50, 56, 0, 56, 56, 0
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OFFSET

1,4


COMMENTS

A square array of n X n pixels can have two states (gray, red). Touching a pixel switches its state and the state of the adjacent pixels. The general problem is to turn all pixels ON given any initial configuration. It requires inverting a n^2 by n^2 matrix in Z/2Z. The sequence is the rank deficiency (corank) of the matrix, such that the zero terms correspond to the sizes for which the general case admits a solution.
The size 5 game can be played at the link given below. Rank deficiency is 2 for that game, but only initial configurations that admit a solution are given.
a(n) is nonzero iff n is in A117870; a(n) is zero iff n is in A076436.  Max Alekseyev, Sep 17 2009
a(n) is even and satisfies a(n) <= n.  Thomas Buchholz, May 19 2014
If a(n) is nonzero, then a(2n+1) is also nonzero.  William Boyles, Jun 28 2018
If n == 5 (modulo 6), then a(n) is nonzero.  William Boyles, Jul 04 2021
For all integers i >= 1 and j >= 0, a((i+1)*j + i) >= a(i).  William Boyles, Aug 04 2021


REFERENCES

See A075462 for references.


LINKS

William Boyles, Table of n, a(n) for n = 1..25000 (first 1000 terms by Max Alekseyev and Thomas Buchholz).
Andries E. Brouwer, Lights Out and Button Madness Games [Gives theory and a(n) for n = 1..1000, Jun 19 2008]
JeuxT45, Turn it red [Dead link]
Martin Kreh, "Lights Out" and Variants, The American Mathematical Monthly 124:10 (2017), 937950.
Eric Weisstein's World of Mathematics, Lights Out Puzzle


FORMULA

Let f(k,x) = U(k,x/2), where U(k,x) is the kth Chebyshev polynomial of the second kind over the field GF(2). So f(0,x)=1, f(1,x)=x, f(2,x)=(1+x)^2, and f(n+1,x)=x*f(n,x)+f(n1,x). Then a(n) equals the degree of gcd(f(n,x), f(n,1+x)). For example, f(5,x)=x^5+x=x(1+x)^4 and f(5, 1+x)=x^4(1+x). So their GCD is x(1+x) and the degree is 2, that is a(5)=2.  Zhao Hui Du, Mar 17 2014; edited by Max Alekseyev, Nov 12 2019


EXAMPLE

For n=2, matrix is [1 1 1 0][1 1 0 1][1 0 1 1][0 1 1 1] which is of full rank.


MATHEMATICA

Table[First[Dimensions[NullSpace[AdjacencyMatrix[GridGraph[{n, n}]] + IdentityMatrix[n*n], Modulus > 2]]], {n, 2, 30}]
(* Or Faster *)
A[k_] := DiagonalMatrix[Array[1 &, k  1], 1] +
DiagonalMatrix[Array[1 &, k  1], 1] + IdentityMatrix[k];
B[k_, 0] := IdentityMatrix[k];
B[k_, 1] := A[k];
B[k_, n_] := B[k, n] = Mod[A[k].B[k, n  1] + B[k, n  2], 2];
Table[First[Dimensions[NullSpace[B[n, n], Modulus > 2]]], {n, 2, 30}]
(* Birkas Gyorgy, Jun 10 2011 *)


PROG

(PARI) { A159257(n) = my(p, q, r); p=Mod(1, 2); q=p*x; for(u=2, n, r=x*q+p; p=q; q=r); p=subst(q, x, 1+x); r=gcd(p, q); poldegree(r) } \\ Zhao Hui Du, Mar 18 2014
(PARI) { A159257(n) = my(f = polchebyshev(n, 2, x/2)*Mod(1, 2)); poldegree( gcd(f, subst(f, x, 1+x)) ); } \\ Max Alekseyev, Nov 12 2019


CROSSREFS

Cf. A075462, A075463, A075464, A076436, A076437.
Sequence in context: A320647 A028572 A107492 * A258997 A232833 A256269
Adjacent sequences: A159254 A159255 A159256 * A159258 A159259 A159260


KEYWORD

nonn


AUTHOR

Bruno Vallet (bruno.vallet(AT)gmail.com), Apr 07 2009


EXTENSIONS

More terms from Max Alekseyev, Sep 17 2009
More terms from Thomas Buchholz, May 16 2014


STATUS

approved



