A135857 Partial sums triangle based on A016777. Riordan convolution triangle ((1 + 2*x)/(1-x)^2, x/(1-x)).

1, 4, 1, 7, 5, 1, 10, 12, 6, 1, 13, 22, 18, 7, 1, 16, 35, 40, 25, 8, 1, 19, 51, 75, 65, 33, 9, 1, 22, 70, 126, 140, 98, 42, 10, 1, 25, 92, 196, 266, 238, 140, 52, 11, 1, 28, 117, 288, 462, 504, 378, 192, 63, 12, 1

Offset: 0

Author

Gary W. Adamson, Dec 01 2007

Keywords

• nonn
• easy
• tabl

Comments

A007318 * a bidiagonal matrix with all 1's in the main diagonal and all 3's in the subdiagonal.

Row sums give A036563(n+2), n >= 0.

From Wolfdieter Lang, Mar 23 2015: (Start)

This is the triangle of iterated partial sums of A016777. Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given).

This is therefore the Riordan triangle ((1+2*x)/(1-x)^2, x/(1-x)) with o.g.f. of the columns ((1+2*x)/(1-x)^2)*(x/(1-x))^k, k >= 0.

The column sequences are A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310, for k = 0..8.

The alternating row sums are A122553(n) = {1, repeat(3)}.

The Riordan A-sequence is A(y) = 1 + y (implying the Pascal triangle recurrence for k >= 1).

The Riordan Z-sequence is A256096, leading to a recurrence for T(n,0) given in the formula section. See the link "Sheffer a- and z-sequences" under A006232 also for Riordan A- and Z-sequences with references. (End)

When the first column (k = 0) is removed from this triangle, the result is A125232. - Georg Fischer, Jul 26 2023

Examples

			The triangle T(n, k) begins:
n\k  0   1   2    3    4    5    6   7   8  9 10 11
0:   1
1:   4   1
2:   7   5   1
3:  10  12   6    1
4:  13  22  18    7    1
5:  16  35  40   25    8    1
6:  19  51  75   65   33    9    1
7:  22  70 126  140   98   42   10   1
8:  25  92 196  266  238  140   52  11   1
9:  28 117 288  462  504  378  192  63  12  1
10: 31 145 405  750  966  882  570 255  75 13  1
11: 34 176 550 1155 1716 1848 1452 825 330 88 14  1
... reformatted and extended by _Wolfdieter Lang_, Mar 23 2015
From _Wolfdieter Lang_, Mar 23 2015: (Start)
T(3, 1) = T(2, 0) + T(2, 1) = 7 + 5 = 12 (Pascal, from the A-sequence given above).
T(4, 0) = 4*T(3, 0) - 9*T(3, 1) + 27*T(3, 2) - 81* T(3, 3) = 4*10 - 9*12 + 27*6 - 81*1 = 13, from the Z-sequence given above and in A256096.
T(4, 0) = 2*T(3, 0) - T(2, 0) = 2*10 - 7 = 13.
(End)
		

Crossrefs

Columns k=0-8 give: A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310.

Cf. A036563, A125232.

Cf. A110813. - Philippe Deléham, Feb 08 2009

Formula

Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal and all 3's in the subdiagonal; i.e., by columns - every column = (1, 3, 0, 0, 0, ...).

T(n,k) = (3n-2k+1)*binomial(n+1,k+1)/(n+1). - Philippe Deléham, Feb 08 2009

From Wolfdieter Lang, Mar 23 2015: (Start)

O.g.f. for row polynomials: (1 + 2*z)/((1- z*(1 + x))*(1 - z)) (see the Riordan property from the comment).

O.g.f. for column k (without leading zeros): (1 + 2*x)/(1-x)^(2+k), k >= 0, (Riordan property).

T(n, k) = T(n-1, k-1) + T(n-1, k) for k >= 1. From the Riordan A-sequence given above in a comment.

T(n, 0) = Sum_{j=0..n} Z(j)*T(n-1, j), for n >= 1, from the Riordan Z-sequence A256096 mentioned above in a comment. Of course, T(n, 0) = 2*T(n-1, 0) - T(n-2, 0) for n >= 2 (see A016777).

(End)

Extensions

Edited. Offset is 0 from the old name and the Philippe Deléham formula. New name, old name as first comment. - Wolfdieter Lang, Mar 23 2015