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

0,3

Comments

The leading diagonal contains 2^n.

Links

Table of n, a(n) for n=0..65.

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

Example

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
  ...

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 *)

Crossrefs

Cf. A112356, A112357, A112359.

Cf. A008949.

Sequence in context: A056242 A343960 A128718 * A126351 A157011 A246173

Adjacent sequences: A112355 A112356 A112357 * A112359 A112360 A112361

Keyword

easy,nonn,tabl

Author

Amarnath Murthy, Sep 05 2005

Extensions

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

Status

approved