OFFSET
1,4
COMMENTS
The matchstick arrangement consists of 3*T(n) = A000217(n) matches. One could also use a card tower with n cards as a basis.
See triangle A085691 for the number of triangles of both orientations.
See the unsigned triangle A122432 with offset 1 for the corresponding case with only the number of triangles oriented like the enclosing triangle.
The first column sequence is A000217(n-1), for n >= 1, and the following ones are then shifted downwards by 2 steps.
Row sums give A002623(n-1), n >= 1, with A002623(-1) = 0. See also the comment of Radu Grigore, Jun 19 2004, in A002623.
The nonzero terms of row 2*k are given in row k-1 of A103217, for k >= 1.
FORMULA
Recurrence: T(n, k) = T(n-1, k) + H(n-2*k+1)*(n-2*k+1), for n >=1, k = 1, 2, ..., n, and T(1, 1) = 0. Here H(x) = 1 for x >= 0 and 0 for x < 0 (a step function)..
T(n, k) = binomial(n-2*k+2, 2), for n >= 1 and k = 1, 2, ..., floor(n/2), and 0 for k = floor(n/2) + 1 .. n. See the comment by Andrew Howroyd in A085691.
EXAMPLE
The triangle T(n, k) begins:
n\k 1 2 3 4 5 6 7 8 9 10 ...
-------------------------------
1: 0
2: 1 0
3: 3 0 0
4 6 1 0 0
5: 10 3 0 0 0
6: 15 6 1 0 0 0
7: 21 10 3 0 0 0 0
8: 28 15 6 1 0 0 0 0
9: 36 21 10 3 0 0 0 0 0
10: 45 28 15 6 1 0 0 0 0 0
...
MATHEMATICA
T[n_, k_]:= If[k<=Floor[n/2], Binomial[n-2*k+2, 2], 0];
Table[T[n, k], {n, 15}, {k, n}]//Flatten (* Amiram Eldar, Apr 23 2020 *)
PROG
(PARI) T(n, k) = if(k <= n\2, binomial(n-2*k+2, 2), 0);
matrxi(10, 10, n, k, T(n, k)) \\ to see the triangle \\ Michel Marcus, May 05 2020
CROSSREFS
KEYWORD
AUTHOR
Wolfdieter Lang, Apr 06 2020
STATUS
approved