A079213 Triangle read by rows giving T(n,k) = number of sets of k black squares in an n X n checkerboard with the upper left corner colored black, such that the line joining any 2 squares slopes down to the right, 0 <= k <= n.

1, 1, 1, 1, 2, 1, 1, 5, 3, 1, 1, 8, 12, 4, 1, 1, 13, 28, 22, 5, 1, 1, 18, 63, 68, 35, 6, 1, 1, 25, 117, 197, 135, 51, 7, 1, 1, 32, 208, 464, 480, 236, 70, 8, 1, 1, 41, 336, 1016, 1376, 996, 378, 92, 9, 1, 1, 50, 525, 2000, 3600, 3372, 1848, 568, 117, 10, 1, 1, 61, 775, 3725

Offset

0,5

Comments

Based on a question from Cees H. Elzinga (ch.elzinga(AT)tiscali.nl), Dec 30 2002

Links

Table of n, a(n) for n=0..69.

Formula

More generally, let f(m, n, k) be the number of such sets in an m X n checkerboard. Then f(m, n, k) = Sum_{k-1<=m'<m, k-1<=n'<n, m+n+m'+n' even} f(m', n', k-1).

G.f.: Sum_{m>=0, n>=0, k>=0} f(m, n, k) x^m * y^n * z^k = (1+x) * (1+y) / ((1-x^2) * (1-y^2) + x*y*z*(1+x*y)).

T(n, 0) = T(n, n) = 1. T(n, 1) = ceiling(n^2/2). T(n, 2) = (n^2 * (n^2-2*n+4))/16 if n is even, ((n-1)^2 * (n^2+3))/16 if n is odd. T(n, n-1) = n. T(n, n-2) = (n-1)*(3n-4)/2.

G.f. (conjectured): Sum_{n>=0, k>=0} T(n, k) x^n y^k = sqrt((1+x)/((1+x-x*y)((1-x)^2 - x*y*(1+x)))).

Conjecture: Sum_{k=0..n} T(n, k) = A025565(n+1).

Example

T(5,3)=22; one of the 22 sets of 3 is shown by the asterisks below; the 'o's denote black squares not in the set.
*.o.o
.*.o.
o.o.*
.o.o.
o.o.o

Mathematica

f[m_, n_, 0] := 1; f[m_, n_, k_] := f[m, n, k]=Sum[If[EvenQ[m+n+mp+np], f[mp, np, k-1], 0], {mp, k-1, m-1}, {np, k-1, n-1}]; T[n_, k_] := f[n, n, k]; Flatten[Table[T[n, k], {n, 0, 11}, {k, 0, n}]]

Crossrefs

Cf. A025565.

Sequence in context: A160364 A107735 A137570 * A047884 A124328 A368487

Adjacent sequences: A079210 A079211 A079212 * A079214 A079215 A079216

Keyword

nonn,easy,tabl

Author

Dean Hickerson, Jan 02 2003

Status

approved