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%I #13 Jan 13 2022 02:32:26
%S 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,
%T 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,
%U 0,1,10,32,52,84,116,48,16,0,0,0,0
%N 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.
%C 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.
%H Pontus von Brömssen, <a href="/A350530/b350530.txt">Antidiagonals n = 1..19, flattened</a>
%e Array begins:
%e n\k| 0 1 2 3 4 5 6 7 8 9 10
%e ---+--------------------------------------------------
%e 1 | 1 1 1 1 1 1 1 1 1 1 1
%e 2 | 0 1 2 3 4 5 6 7 8 9 10
%e 3 | 0 0 0 2 4 8 12 18 24 32 40
%e 4 | 0 0 0 0 0 4 12 28 52 84 132
%e 5 | 0 0 0 0 0 8 36 84 176 332 568
%e 6 | 0 0 0 0 4 28 116 308 704 1396 2548
%e 7 | 0 0 0 0 4 48 232 728 2104 4940 11008
%e 8 | 0 0 0 0 16 100 556 1936 7092 19908 49364
%e 9 | 0 0 0 0 12 176 1348 6588 23356 74228 202504
%e 10 | 0 0 0 0 8 268 2492 15544 72820 259800 842688
%e 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.
%o (Python)
%o def A350530_col(k,nmax):
%o d = []
%o c = [0]*nmax
%o while 1:
%o if not d or all(d[-1][:-1]):
%o if d and d[-1][-1] == 0:
%o c[len(d)-1] += 1 + (0 != 2*d[0][0] != k+1)
%o elif len(d) < nmax:
%o d.append([-1])
%o for i in range(len(d)-1):
%o d[-1].append(d[-1][-1]-d[-2][i])
%o while d and d[-1][0] == k:
%o d.pop()
%o if not d or len(d) == 1 and 2*d[0][0] >= k: return c
%o for i in range(len(d)):
%o d[-1][i] += 1
%Y Cf. A200154, A350365, A350529.
%Y Rows: A000012 (n=1), A001477 (n=2), A007590 (n=3).
%Y Columns: A000007 (k=0), A019590 (k=1), A130706 (k=2).
%K nonn,tabl
%O 1,8
%A _Pontus von Brömssen_, Jan 03 2022
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