0,5

Rows 0 and 2 are included by extension since they fit the formula. Row 1 equals the odd numbers in order that triangle A108556 maintains that A108556(n,n-1) = (n/2)*A108556(n,n) for all n>=1, where row n of triangle A108556 equals the inverse binomial transform of row n of this square array.

T(n, k) = Sum_{j=0..n} C(n+k-j, k-j)*[C(2*n, 2*j) - 2*j*(n-j)*C(n, j)/(n-1)] for n>1, with T(0, k)=1, T(1, k)=2*k+1.

Square array begins:

1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...

1,3,5,7,9,11,13,15,17,19,21,23,25,27,...

1,5,13,25,41,61,85,113,145,181,221,265,...

1,13,55,147,309,561,923,1415,2057,2869,...

1,25,169,625,1681,3721,7225,12769,21025,...

1,41,411,2051,6981,18733,42783,86983,...

1,61,853,5577,23673,76389,204205,476113,...

1,85,1583,13203,68853,264825,824083,...

Inverse binomial transform of rows gives

rows of triangle A108556:

1,

1,2,

1,4,4,

1,12,30,20,

1,24,120,192,96,

1,40,330,940,1080,432, ...

Product of the g.f. of row n and (1-x)^(n+1)

generates the symmetric triangle A108558:

1;

1,1;

1,2,1;

1,9,9,1;

1,20,54,20,1;

1,35,180,180,35,1; ...

The row sums of triangle A108558 equals the

main diagonal of triangle A108556.

(PARI) {T(n, k)=if(n<0|k<0, 0, if(n==0|k==0, 1, if(n==1, 2*k+1, sum(j=0, k, binomial(n+k-j, k-j)*(binomial(2*n, 2*j)-2*n*binomial(n-2, j-1))))))}

Cf. A108554 (diagonal), A108555 (antidiagonal sums), A108556, A108558, A001844 (row 2), A005902 (row 3), A007204 (row 4), A008356 (row 5), A008358 (row 6), A008360 (row 7), A008362 (row 8), A008377 (row 9), A008379 (row 10).

Sequence in context: A137897 A056152 A125690 this_sequence A158418 A124925 A073145

Adjacent sequences: A108550 A108551 A108552 this_sequence A108554 A108555 A108556

nonn,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Jun 10 2005