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%I #24 Jan 19 2025 17:11:55
%S 1,2,5,3,8,7,4,12,11,10,6,17,16,15,14,9,25,23,22,21,18,13,37,34,32,31,
%T 27,20,19,54,50,47,45,40,30,24,28,79,73,69,66,58,44,36,26,41,116,107,
%U 101,97,85,64,53,39,29,60,170,157,148,142,125,94,77,57,43,33,88,249,230
%N The 3rd-order Zeckendorf array, T(n,k), read by antidiagonals.
%C Rows satisfy this recurrence: T(n,k) = T(n,k-1) + T(n,k-3) for all k>=4.
%C Except for initial terms, (row 1) = A000930 (column 1) = A020942 (column 2) = A064105 (column 3) = A064106.
%C As a sequence, the array is a permutation of the natural numbers.
%C As an array, T is an interspersion (hence also a dispersion).
%H Clark Kimberling, <a href="https://www.fq.math.ca/Scanned/33-1/kimberling.pdf">The Zeckendorf array equals the Wythoff array</a>, Fibonacci Quarterly 33 (1995) 3-8.
%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%F Row 1 is the 3rd-order Zeckendorf basis, given by initial terms b(1)=1, b(2)=2, b(3)=3 and recurrence b(k) = b(k-1) + b(k-3) for k>=4. Every positive integer has a unique 3-Zeckendorf representation: n = b(i(1)) + b(i(2)) + ... + b(i(p)), where |i(h)-i(j))>=3. Rows of T are defined inductively: T(n,1) is the least positive integer not in an earlier row. T(n,2) is obtained from T(n,1) as follows: if T(n,1) = b(i(1)) + b(i(2)) + ... + b(i(p)), then T(n,k+1) = b(i(1+k)) + b(i(2+k)) + ... + b(i(p+k)) for k=1,2,3,... .
%F A(n, k) = A000930(k)*A202342(n) + A000930(k-2)*A136495(n) + A000930(k-1)*(n-1) for n > 1. - _Alan Michael Gómez Calderón_, Dec 23 2024
%e Northwest corner:
%e 1 2 3 4 6 9 13 19 ...
%e 5 8 12 17 25 37 54 79 ...
%e 7 11 16 23 34 50 73 107 ...
%e 10 15 22 32 47 69 101 148 ...
%e ...
%Y Cf. A000930, A035513, A134563, A136175, A136495, A202342.
%K nonn,tabl,changed
%O 1,2
%A _Clark Kimberling_, Dec 20 2007