A112358 The following triangle is based on Pascal's triangle. The r-th term of the n-th row is sum of C(n,r) successive integers so that the sum of all the terms of the row is (2^n)*(2^n+1)/2, the 2^n -th triangular number. Sequence contains the triangle read by rows.

1, 1, 2, 1, 5, 4, 1, 9, 18, 8, 1, 14, 51, 54, 16, 1, 20, 115, 215, 145, 32, 1, 27, 225, 650, 750, 363, 64, 1, 35, 399, 1645, 2870, 2310, 868, 128, 1, 44, 658, 3668, 8995, 10724, 6538, 2012, 256, 1, 54, 1026, 7434, 24381, 40257, 35658, 17442, 4563, 512, 1, 65, 1530, 13980, 59115, 129150, 156135, 109020, 44595, 10185, 1024

Offset: 0

Author

Amarnath Murthy, Sep 05 2005

Keywords

• easy
• nonn
• tabl

Comments

The leading diagonal contains 2^n.

Examples

			Row for n = 3 is 1, (2+3+4), (5+6+7), 8.
Triangle begins:
  1
  1 2
  1 5 4
  1 9 18 8
  1 14 51 54 16
  ...
		

Crossrefs

Cf. A112356, A112357, A112359.

Cf. A008949.

Programs

Mathematica

A008949[n_, k_] := Sum[Binomial[n, j], {j, 0, k}];
A112358[n_, k_] := If[k == 0, 1, Binomial[A008949[n, k] + 1, 2] - Binomial[A008949[n, k - 1] + 1, 2]];
Table[A112358[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 01 2023 *)

Formula

T(n,0) = 1, T(n,k) = C(A008949(n,k)+1, 2) - C(A008949(n,k-1)+1, 2) = C(n,k)*(A008949(n+1,k)+1)/2 for k>0. - Franklin T. Adams-Watters, Sep 27 2006

Extensions

More terms from Amber Reardon (alr5041(AT)psu.edu) and Vincent M. DelPrince (vmd5003(AT)psu.edu), Oct 04 2005