A106255 Triangle composed of triangular numbers, row sums = A006918.

1, 1, 1, 1, 3, 1, 1, 3, 3, 1, 1, 3, 6, 3, 1, 1, 3, 6, 6, 3, 1, 1, 3, 6, 10, 6, 3, 1, 1, 3, 6, 10, 10, 6, 3, 1, 1, 3, 6, 10, 15, 10, 6, 3, 1, 1, 3, 6, 10, 15, 15, 10, 6, 3, 1, 1, 3, 6, 10, 15, 21, 15, 10, 6, 3, 1

Offset: 1

Author

Gary W. Adamson, Apr 28 2005

Keywords

• nonn
• tabl

Comments

Perform the operation Q * R; Q = infinite lower triangular matrix with 1, 2, 3, ... in each column (offset, fill in spaces with zeros). Q = upper right triangular matrix of the form:

1, 1, 1, 1, ...

0, 1, 1, 1, ...

0, 0, 1, 1, ...

0, 0, 0, 1, ...

Q * R generates an array:

1, 1, 1, 1, ...

1, 3, 3, 3, ...

1, 3, 6, 6, ...

1, 3, 6, 10, ...

...

... from which we take antidiagonals forming the rows of this triangle.

Examples

			From _Boris Putievskiy_, Jan 13 2013: (Start)
The start of the sequence as table:
  1, 1, 1,  1,  1,  1, ...
  1, 3, 3,  3,  3,  3, ...
  1, 3, 6,  6,  6,  6, ...
  1, 3, 6, 10, 10, 10, ...
  1, 3, 6, 10, 15, 15, ...
  1, 3, 6, 10, 15, 21, ...
  1, 3, 6, 10, 15, 21, ...
  ...
(End)
Triangle rows or columns can be generated by following the triangle format:
  1;
  1, 1;
  1, 3, 1;
  1, 3, 3,  1;
  1, 3, 6,  3,  1;
  1, 3, 6,  6,  3,  1;
  1, 3, 6, 10,  6,  3,  1;
  1, 3, 6, 10, 10,  6,  3,  1;
  1, 3, 6, 10, 15, 10,  6,  3, 1;
  1, 3, 6, 10, 15, 15, 10,  6, 3, 1;
  1, 3, 6, 10, 15, 21, 15, 10, 6, 3, 1;
  ...
		

Links

• Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.

Crossrefs

Cf. A006918 (row sums, without the zero), A002260, A004736.

Programs

Mathematica

p[x_, n_] = Sum[x^i*If[i ==Floor[n/2] && Mod[n, 2] == 0, 0, If[i <= Floor[n/2], 2*(i + 1), -(2*((n + 1) - i))]], {i, 0, n}]/(2*(1 - x));
Table[CoefficientList[FullSimplify[p[x, n]], x], {n, 1, 11}];
Flatten[%]

Formula

From Boris Putievskiy, Jan 13 2013: (Start)

T(n,k) = min(n*(n+1)/2,k*(k+1)/2), read by antidiagonals.

a(n) = min(A002260(n)*(A002260(n)+1)/2, A004737(n)*(A004737(n)+1)/2).

a(n) = min(i*(i+1)/2, j*(j+1)/2), where

i = n-t*(t+1)/2,

j = (t*t+3*t+4)/2-n,

t = floor((-1+sqrt(8*n-7))/2). (End)

Extensions

Additional comments from Roger L. Bagula and Gary W. Adamson, Apr 02 2009