A385436 Tribonacci array of the second kind, read by upward antidiagonals.
0, 2, 1, 4, 5, 3, 6, 8, 10, 7, 9, 12, 16, 20, 14, 11, 18, 23, 31, 38, 27, 13, 21, 34, 44, 58, 71, 51, 15, 25, 40, 64, 82, 108, 132, 95, 17, 29, 47, 75, 119, 152, 200, 244, 176, 19, 32, 54, 88, 139, 220, 281, 369, 450, 325, 22, 36, 60, 101, 163, 257, 406, 518, 680
Offset: 1
Examples
Array including some prepended columns (p = 1..4): p=4 p=3 p=2 p=1 | k=1 k=2 k=3 k=4 k=5 k=6 k=7 k=8 k=9 k=10 -2 -1 -1 -1 | 0 1 3 7 14 27 51 95 176 325 -2 -1 0 0 | 2 5 10 20 38 71 132 244 450 829 -2 0 0 1 | 4 8 16 31 58 108 200 369 680 -2 0 1 2 | 6 12 23 44 82 152 281 518 -1 0 2 4 | 9 18 34 64 119 220 406 748 -1 1 2 5 | 11 21 40 75 139 257 474 873 -1 1 3 6 | 13 25 47 88 163 301 555 1022 -1 1 4 7 | 15 29 54 101 187 345 636 1171 -1 2 4 8 | 17 32 60 112 207 382 704 1296 -1 2 5 9 | 19 36 67 125 0 2 6 11 | 22 42 78 145 Each row of the array satisfies the recurrence relation T(m,k) = 2*T(m,k-1) - T(m,k-4); from this, the prepended columns are obtained by rowwise backward recursion.
Links
- A.H.M. Smeets, Table of n, a(n) for n = 1..20100 (first 200 antidiagonals).
- Larry Ericksen and Peter G. Anderson, Patterns in differences between rows in k-Zeckendorf arrays, The Fibonacci Quarterly, Vol. 50, No. 1 (February 2012), pp. 11-18.
- Clark Kimberling, The Zeckendorf array equals the Wythoff array, Fibonacci Quarterly 33 (1995) 3-8.
- Eric Weisstein's World of Mathematics, Interspersion.
- Eric Weisstein's World of Mathematics, Sequence Dispersion.
Crossrefs
Programs
-
Python
def ToDual_111_Zeck(n): if n == 0: return "0" f0, f1, f2, sf = 1, 0, 0, 0 while n > sf: f0, f1, f2 = f0+f1+f2, f0, f1 sf += f0 r, s = sf-n, "1" while f0 > 1: f0, f1, f2 = f1, f2, f0-f1-f2 r, s = r%f0, s+str(1-r//f0) return s def From_111_Zeck(s): f0, f1, f2, i, n = 1, 1, 0, len(s), 0 while i > 0: i -= 1 f0, f1, f2, n = f0+f1+f2, f0, f1, n+int(s[i])*f0 return n d, a, n, c1 = 0, 0, 0, [] while d < 11: s = ToDual_111_Zeck(a) if s[len(s)-1] == "0": # == even n, d = n+1, d+1 print(a, end = ", ") i, c1, p1 = d-1, c1+[s], "" while i > 0: n, i, p1 = n+1, i-1, p1+"1" print(From_111_Zeck(c1[i]+p1), end = ", ") a += 1
Comments