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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.

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