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%I #4 Nov 11 2015 19:43:33
%S 4,13,8,49,55,16,181,490,233,32,676,3567,4900,987,64,2521,28925,70669,
%T 49000,4181,128,9409,223356,1243225,1399783,490000,17711,256,35113,
%U 1759250,20386617,53429620,27726581,4900000,75025,512,131044,13750304
%N 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 +-(.,.) 0,0 0,1 or 1,2.
%C Table starts
%C ....4......13.........49...........181..............676.................2521
%C ....8......55........490..........3567............28925...............223356
%C ...16.....233.......4900.........70669..........1243225.............20386617
%C ...32.....987......49000.......1399783.........53429620...........1855980772
%C ...64....4181.....490000......27726581.......2296230561.........168990466353
%C ..128...17711....4900000.....549201567......98684484373.......15386771913704
%C ..256...75025...49000000...10878455069....4241136597604.....1400983500645217
%C ..512..317811..490000000..215477871383..182270189212469...127561175981852920
%C .1024.1346269.4900000000.4268134837381.7833376999538689.11614593343457551705
%H R. H. Hardin, <a href="/A264341/b264341.txt">Table of n, a(n) for n = 1..143</a>
%F Empirical for column k:
%F k=1: a(n) = 2*a(n-1)
%F k=2: a(n) = 4*a(n-1) +a(n-2)
%F k=3: a(n) = 10*a(n-1)
%F k=4: a(n) = 19*a(n-1) +16*a(n-2)
%F k=5: a(n) = 43*a(n-1) -43*a(n-3) +a(n-4)
%F k=6: a(n) = 87*a(n-1) +374*a(n-2) -470*a(n-3) +207*a(n-4) +3*a(n-5)
%F k=7: a(n) = 191*a(n-1) +1102*a(n-2) -7594*a(n-3) -38349*a(n-4) +38507*a(n-5)
%F Empirical for row n:
%F n=1: a(n) = 4*a(n-1) -4*a(n-3) +a(n-4)
%F n=2: [order 14]
%F n=3: [order 34]
%e Some solutions for n=3 k=4
%e ..7..8..9..3..4....1..0..3..2..4....7..8..2..3..4....1..2..9..4..3
%e .12..5..0..1..2...12..6..7..9..8....5..6..0..1..9...12..6..0..7..8
%e .17.10.13..6.14...11.10..5.14.13...11.10.12.13.14...10.18..5.14.13
%e .15.16.18.11.19...16.15.17.18.19...15.17.16.18.19...15.17.16.11.19
%Y Column 1 is A000079(n+1).
%Y Column 2 is A033887(n+1).
%Y Row 1 is A097948.
%K nonn,tabl
%O 1,1
%A _R. H. Hardin_, Nov 11 2015
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