OFFSET
0,2
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
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)
EXAMPLE
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
KEYWORD
AUTHOR
Gary W. Adamson, Dec 01 2007
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
STATUS
approved