0,2

G.f. of row n: Sum_{k=0..n} (x^binomial(n,k) - 1)/(x-1) = Sum_{k=0..binomial(n,n\2)-1} T(n,k)*x^k.

A152548(n) = Sum_{k=0..C(n,[n/2])-1} T(n,k)^2 = Sum_{k=0..[(n+1)/2]} C(n+1, k)*(n+1-2k)^3/(n+1).

The number of terms in row n is C(n,[n/2]).

Triangle begins:

[1],

[2],

[3,1],

[4,2,2],

[5,3,3,3,1,1],

[6,4,4,4,4,2,2,2,2,2],

[7,5,5,5,5,5,3,3,3,3,3,3,3,3,3,1,1,1,1,1],

[8,6,6,6,6,6,6,4,4,4,4,4,4,4,4,4,4,4,4,4,4,2,2,2,2,2,2,2,2,2,2,2,2,2,2],

[9,7,7,7,7,7,7,7,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1],

...

ILLUSTRATION OF GENERATING METHOD.

Row n is derived from the binomial coefficients in the following way.

Place markers in an array so that the number of contiguous markers

in row k is C(n,k) and then count the markers along columns.

For example, row 6 of this triangle is generated from C(6,k) like so:

------------------------------------------

1: o - - - - - - - - - - - - - - - - - - -

6: o o o o o o - - - - - - - - - - - - - -

15:o o o o o o o o o o o o o o o - - - - -

20:o o o o o o o o o o o o o o o o o o o o

15:o o o o o o o o o o o o o o o - - - - -

6: o o o o o o - - - - - - - - - - - - - -

1: o - - - - - - - - - - - - - - - - - - -

------------------------------------------

Counting the markers along the columns gives row 6 of this triangle:

[7,5,5,5,5,5,3,3,3,3,3,3,3,3,3,1,1,1,1,1].

Continuing in this way generates all the rows of this triangle.

...

Number of repeated terms in each row of this triangle forms A008315:

1;

1;

1, 1;

1, 2;

1, 3, 2;

1, 4, 5;

1, 5, 9, 5;

1, 6, 14, 14;

1, 7, 20, 28, 14;...

(PARI) {T(n, k)=polcoeff(sum(j=0, n, (x^binomial(n, j) - 1)/(x-1)), k)}

Cf. A152548 (row squared sums), A008315; A152545.

Sequence in context: A087088 A104705 A143361 this_sequence A083906 A160541 A022446

Adjacent sequences: A152544 A152545 A152546 this_sequence A152548 A152549 A152550

nonn,tabf

Paul D. Hanna (pauldhanna(AT)juno.com), Dec 14 2008