A350529 Square array read by antidiagonals downwards: T(n,k) is the number of sequences of length n with terms in 1..k such that no iterated difference is zero, n, k >= 0.
1, 1, 0, 1, 1, 0, 1, 2, 0, 0, 1, 3, 2, 0, 0, 1, 4, 6, 2, 0, 0, 1, 5, 12, 10, 2, 0, 0, 1, 6, 20, 32, 16, 2, 0, 0, 1, 7, 30, 72, 86, 26, 2, 0, 0, 1, 8, 42, 138, 256, 232, 42, 2, 0, 0, 1, 9, 56, 234, 624, 906, 622, 68, 2, 0, 0
Offset: 0
Examples
n\k| 0 1 2 3 4 5 6 7 8 9 10 ---+-------------------------------------------------------------------------- 0 | 1 1 1 1 1 1 1 1 1 1 1 1 | 0 1 2 3 4 5 6 7 8 9 10 2 | 0 0 2 6 12 20 30 42 56 72 90 3 | 0 0 2 10 32 72 138 234 368 544 770 4 | 0 0 2 16 86 256 624 1278 2370 4030 6462 5 | 0 0 2 26 232 906 2790 6900 15096 29536 53678 6 | 0 0 2 42 622 3180 12366 36964 95494 215146 443464 7 | 0 0 2 68 1662 11116 54572 197294 601986 1562274 3652850 8 | 0 0 2 110 4426 38754 240278 1051298 3788268 11325490 30041458 9 | 0 0 2 178 11774 134902 1056546 5595236 23814458 82024662 246853482 10 | 0 0 2 288 31316 469306 4643300 29762654 149631992 593798912 2027577296 For n = 4 and k = 3, the following T(4,3) = 16 sequences are counted: 1212, 1213, 1312, 1313, 1323, 2121, 2131, 2132, 2312, 2313, 2323, 3121, 3131, 3132, 3231, 3232.
Links
- Pontus von Brömssen, Antidiagonals n = 0..20, flattened
Crossrefs
Programs
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Python
def A350529_col(k,nmax): d = [] c = [0]*(nmax+1) while 1: if not d or all(d[-1]): c[len(d)] += 1 + (bool(d) and 2*d[0][0] != k+1) if len(d) < nmax: d.append([0]) 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
Comments