A350530 Square array read by antidiagonals downwards: T(n,k) is the number of sequences of length n with terms in 0..k such that the (n-1)-st difference is zero, but no earlier iterated difference is zero, n, k >= 1.

1, 1, 0, 1, 1, 0, 1, 2, 0, 0, 1, 3, 0, 0, 0, 1, 4, 2, 0, 0, 0, 1, 5, 4, 0, 0, 0, 0, 1, 6, 8, 0, 0, 0, 0, 0, 1, 7, 12, 4, 0, 0, 0, 0, 0, 1, 8, 18, 12, 8, 4, 0, 0, 0, 0, 1, 9, 24, 28, 36, 28, 4, 0, 0, 0, 0, 1, 10, 32, 52, 84, 116, 48, 16, 0, 0, 0, 0

Offset: 1

Author

Pontus von Brömssen, Jan 03 2022

Keywords

• nonn
• tabl

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.

Examples

			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.
		

Links

• Pontus von Brömssen, Antidiagonals n = 1..19, flattened

Crossrefs

Cf. A200154, A350365, A350529.

Rows: A000012 (n=1), A001477 (n=2), A007590 (n=3).

Columns: A000007 (k=0), A019590 (k=1), A130706 (k=2).

Programs

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