%I #14 Jun 17 2021 05:53:38
%S 1,2,6,3,9,8,4,13,12,11,5,18,17,16,15,7,24,23,22,21,20,10,33,31,30,29,
%T 28,25,14,46,43,41,40,39,35,27,19,64,60,57,55,54,49,38,32,26,88,83,79,
%U 76,74,68,53,45,34,36,121,114,109,105,102,93,73,63,48,37,50,167,157,150
%N The 4th-order Zeckendorf array, T(n,k), read by antidiagonals.
%C Rows satisfy this recurrence: T(n,k) = T(n,k-1) + T(n,k-4) for all k>=5.
%C Except for initial terms, (row 1) = A003269 (row 2) = A014101.
%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="http://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 4th-order Zeckendorf basis, given by initial terms b(1)=1, b(2)=2, b(3)=3, b(4)=4 and recurrence b(k) = b(k-1) + b(k-4) for k>=5. Every positive integer has a unique 4-Zeckendorf representation: n = b(i(1)) + b(i(2)) + ... + b(i(p)), where |i(h) - i(j)| >= 4. 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,... .
%e Northwest corner:
%e 1 2 3 4 5 7 10 14 ...
%e 6 9 13 18 24 33 46 64 ...
%e 8 12 17 23 31 43 60 83 ...
%e 11 16 22 30 41 57 79 109 ...
%Y Cf. A003269 (row n=1), A134564.
%K nonn,tabl
%O 1,2
%A _Clark Kimberling_, Dec 20 2007