A224838 Triangle read by rows, obtained from triangle A011973 by reading that array from right to left along the irregular paths shown in the figure.

1, 1, 1, 2, 1, 1, 3, 1, 1, 3, 4, 1, 4, 6, 5, 1, 1, 10, 10, 6, 1, 1, 5, 20, 15, 7, 1, 6, 15, 35, 21, 8, 1, 1, 21, 35, 56, 28, 9, 1, 1, 7, 56, 70, 84, 36, 10, 1, 8, 28, 126, 126, 120, 45, 11, 1, 1, 36, 84, 252, 210, 165, 55, 12, 1, 1, 9, 120, 210, 462, 330, 220, 66, 13, 1

Offset: 1

Author

John Molokach, Jul 21 2013

Keywords

• nonn
• tabf

Comments

The successive rows have lengths 1,2,2; 3,4,4; 5,6,6; 7,8,8; ...

Sum of row n is A005314(n).

Old definition was: "Triangle of falling diagonals of A011973 (with rows displayed as centered text)."

Examples

			First 11 rows of the triangle:
  1;
  1,  1;
  2,  1;
  1,  3,  1;
  1,  3,  4,  1;
  4,  6,  5,  1;
  1, 10, 10,  6,  1;
  1,  5, 20, 15,  7,  1;
  6, 15, 35, 21,  8,  1;
  1, 21, 35, 56, 28,  9,  1;
  1,  7, 56, 70, 84, 36, 10,  1;
		

Links

• N. J. A. Sloane, Construction of present triangle by reading triangle A011973 from right to left along the paths indicated.

Crossrefs

Cf. A005314, A227300, A001973, A000045, A004396.

Programs

Mathematica

Table[Reverse[Table[Binomial[n - Floor[(k + 1)/2], n - Floor[(3 k - 1)/2]], {k, Floor[(2 n + 2)/3]}]], {n, 13}] (* T. D. Noe, Jul 25 2013 *)
Column[Table[Binomial[n - Floor[(4 n + 15 - 6 k + (-1)^k)/12], n - Floor[(4 n + 15 - 6 k + (-1)^k)/12] - Floor[(2 n - 1)/3] + k - 1], {n, 1, 25}, {k, 1, Floor[(2 n + 2)/3]}]] (* John Molokach, Jul 25 2013 *)

Formula

r(n) = binomial(n-floor((4n+15-6k+(-1)^k)/12), n-floor((4n+15-6k+(-1)^k)/12)-floor((2n-1)/3)+k-1), k = 1..floor((2n+2)/3).

R(n) = binomial(n-floor((k+1)/2), n-floor((3k-1)/2)), k = 1..floor((2n+2)/3), gives the terms of each row in reverse order.

Extensions

Entry revised by N. J. A. Sloane, Jul 07 2024