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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 (list; graph; refs; listen; history; text; internal format)
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 consists in turning 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

REFERENCES

See A075462 for references.

LINKS

Max Alekseyev and Thomas Buchholz, Table of n, a(n) for n = 1..1000 [recomputed these a(n), May 16 2014]

Andries E. Brouwer, Lights Out and Button Madness Games [Gives theory and a(n) for n = 1..1000, Jun 19 2008]

Turn it red

Eric Weisstein's World of Mathematics, Lights Out Puzzle

FORMULA

Let f(k, 2x) be the Chebyshev Polynomial of the Second Kind in Limit Field F_2. So f(0,x)=1, f(1,x)=x, f(2,x)=(1+x)^2, f(n+1,x)=x*f(n,x)+f(n-1,x). The degree of gcd(f(n,x), f(n, 1+x)) is this sequence. For example, f(5,x)=x^5+x=x(1+x)^4. f(5, 1+x)=x^4(1+x). So their GCD is x(1+x) and the degree is 2. So the 5th element of the array is 2. - Zhao Hui Du, Mar 17 2014

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

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

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Last modified July 20 16:15 EDT 2019. Contains 325185 sequences. (Running on oeis4.)