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A263973 T(n,k)=Number of (n+1)X(k+1) arrays of permutations of 0..(n+1)*(k+1)-1 with each element having index change (+-,+-) 0,2 1,1 or 1,0.

%I A263973 #6 Nov 10 2015 10:30:19
%S A263973 4,20,0,108,72,16,465,1817,1441,0,2265,10720,86633,13856,64,10920,
%T A263973 89472,2604481,2940624,189993,0,52752,1791696,102742049,226810352,
%U A263973 118376593,2215960,256,253176,27783305,4952396901,24130893568,31702909989
%N A263973 T(n,k)=Number of (n+1)X(k+1) arrays of permutations of 0..(n+1)*(k+1)-1 with each element having index change (+-,+-) 0,2 1,1 or 1,0.
%C A263973 Table starts
%C A263973 ....4........20..........108...........465...........2265.........10920
%C A263973 ....0........72.........1817.........10720..........89472.......1791696
%C A263973 ...16......1441........86633.......2604481......102742049....4952396901
%C A263973 ....0.....13856......2940624.....226810352....24130893568.4071813326144
%C A263973 ...64....189993....118376593...31702909989.11743850653769
%C A263973 ....0...2215960...4465927225.3594130469760
%C A263973 ..256..28027420.173624682220
%C A263973 ....0.340646784
%C A263973 .1024
%H A263973 R. H. Hardin, <a href="/A263973/b263973.txt">Table of n, a(n) for n = 1..49</a>
%F A263973 Empirical for column k:
%F A263973 k=1: a(n) = 4*a(n-2)
%F A263973 k=2: [order 10]
%F A263973 k=3: [order 27]
%F A263973 k=4: [order 53]
%F A263973 Empirical for row n:
%F A263973 n=1: [linear recurrence of order 22]
%e A263973 Some solutions for n=3 k=4
%e A263973 ..2..3..6..9..8....2..3..6..1..9....2..5..4..1..8....2..5..4..9..8
%e A263973 ..7..0..1..4.14....7..0..5.12..4....7..0..3.14.13....7..0..1.13..3
%e A263973 ..5.15.17.18.19...16.15..8.17.18....6.16.18.19..9...12..6.17.19.18
%e A263973 .10.12.11.16.13...10.11.19.13.14...11.10.15.12.17...11.10.15.16.14
%Y A263973 Column 1 is A199572(n+1).
%Y A263973 Row 1 is A263965.
%K A263973 nonn,tabl
%O A263973 1,1
%A A263973 _R. H. Hardin_, Oct 30 2015