OFFSET
1,8
COMMENTS
For fixed n, T(n,k) is a quasi-polynomial of degree n-1 in k. For example, T(4,k) = (8/27)*k^3 - 2*k^2 + b(k)*k + c(k), where b and c are periodic with period 3.
LINKS
Pontus von Brömssen, Antidiagonals n = 1..19, flattened
EXAMPLE
Array begins:
n\k| 0 1 2 3 4 5 6 7 8 9 10
---+--------------------------------------------------
1 | 1 1 1 1 1 1 1 1 1 1 1
2 | 0 1 2 3 4 5 6 7 8 9 10
3 | 0 0 0 2 4 8 12 18 24 32 40
4 | 0 0 0 0 0 4 12 28 52 84 132
5 | 0 0 0 0 0 8 36 84 176 332 568
6 | 0 0 0 0 4 28 116 308 704 1396 2548
7 | 0 0 0 0 4 48 232 728 2104 4940 11008
8 | 0 0 0 0 16 100 556 1936 7092 19908 49364
9 | 0 0 0 0 12 176 1348 6588 23356 74228 202504
10 | 0 0 0 0 8 268 2492 15544 72820 259800 842688
For n = 4 and k = 6, the following T(4,6) = 12 sequences are counted: 1454, 1564, 2125, 2565, 3126, 3236, 4541, 4651, 5212, 5652, 6213, 6323.
PROG
(Python)
def A350530_col(k, nmax):
d = []
c = [0]*nmax
while 1:
if not d or all(d[-1][:-1]):
if d and d[-1][-1] == 0:
c[len(d)-1] += 1 + (0 != 2*d[0][0] != k+1)
elif len(d) < nmax:
d.append([-1])
for i in range(len(d)-1):
d[-1].append(d[-1][-1]-d[-2][i])
while d and d[-1][0] == k:
d.pop()
if not d or len(d) == 1 and 2*d[0][0] >= k: return c
for i in range(len(d)):
d[-1][i] += 1
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Pontus von Brömssen, Jan 03 2022
STATUS
approved