1,5

To be an acceptable row, there must be at least one run of white squares and all runs of white squares must be of length at least three. Palindromic rows are of interest since if n is odd, the middle row of an n x n crossword puzzle must be palindromic if the puzzle is to have to usual rotational symmetry. Rather than use the explicit formula above, it is perhaps easier to observe that for all m, a(2m) = a(2m-1), and thus both the subsequences of odd-numbered terms and even-numbered terms are Fibonacci numbers - 1. (see A000071)

Recurrence: a[n + 4] = a[n + 2] + a[n] + 1, a[1] = 0, a[2] = 0, a[3] = 1, a[4] = 1

a(n) = (2^(-n/2) (-5*2^(n/2)*(-1 + sqrt(5))^(3/2) + sqrt(5)*(1 + sqrt(5))^(n/2) (-sqrt(2)*(-1 + (-1)^n) + (1 + (-1)^n) sqrt(-1 + sqrt(5))) + 2 (-1 + sqrt(5))^(n/2) (sqrt(-55 + 25 sqrt(5)) cos(n Pi)/2 + sqrt(2) (-5 + 2 sqrt(5)) sin(n Pi)/2))))/(5 (-1 + sqrt(5))^(3/2)).

O.g.f.: x^3/((-1+x)*(-1+x^2+x^4)). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Dec 05 2007

a(9) = 7 because the palindromic rows, using 0's for white squares and 1's for black, are: 000000000, 100000001, 110000011, 111000111, 000010000, 000111000, 100010001

a[n_] := FullSimplify[(-5*2^(n/2)*(-1 + Sqrt[5])^(3/2) + Sqrt[5]*(1 + Sqrt[5])^(n/2)*((-Sqrt[2])*(-1 + (-1)^n) + (1 + (-1)^n)*Sqrt[ -1 + Sqrt[5]]) + 2*(-1 + Sqrt[5])^(n/2)*(Sqrt[ -55 + 25*Sqrt[5]]*Cos[(n*Pi)/2] + Sqrt[2]*(-5 + 2*Sqrt[5])*Sin[(n*Pi)/2]))/(2^(n/2)*(5*(-1 + Sqrt[5])^(3/2)) )]; Array[ n, 51] - Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 28 2007

Cf. A000071, A130578.

Adjacent sequences: A131521 A131522 A131523 this_sequence A131525 A131526 A131527

Sequence in context: A064410 A062896 A025065 this_sequence A089075 A011142 A060029

nonn

Marc Brodie (mbrodie(AT)wju.edu), Aug 24 2007

More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 28 2007