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A250387 T(n,k)=Number of length n+3 0..k arrays with no four consecutive terms having the maximum of any two terms equal to the minimum of the remaining two terms 13

%I #4 Nov 20 2014 08:47:31

%S 6,42,6,156,78,6,420,432,146,6,930,1560,1208,274,6,1806,4350,5848,

%T 3384,514,6,3192,10206,20518,21950,9480,966,6,5256,21168,58114,96866,

%U 82398,26578,1816,6,8190,40032,141344,331128,457366,309452,74528,3414,6,12210

%N T(n,k)=Number of length n+3 0..k arrays with no four consecutive terms having the maximum of any two terms equal to the minimum of the remaining two terms

%C Table starts

%C .6....42.....156......420........930........1806........3192.........5256

%C .6....78.....432.....1560.......4350.......10206.......21168........40032

%C .6...146....1208.....5848......20518.......58114......141344.......306816

%C .6...274....3384....21950......96866......331128......944272......2352468

%C .6...514....9480....82398.....457366.....1886948.....6308960.....18038628

%C .6...966...26578...309452....2160160....10755158....42158796....138336744

%C .6..1816...74528..1162292...10203222....61304844...281731072...1060924200

%C .6..3414..208998..4365720...48194820...349446100..1882722300...8136444312

%C .6..6418..586102.16398414..227649458..1991900312.12581692558..62400176438

%C .6.12066.1643650.61595866.1075313612.11354195442.84080003928.478561255176

%H R. H. Hardin, <a href="/A250387/b250387.txt">Table of n, a(n) for n = 1..9999</a>

%F Empirical for column k, apparently recurrence of order 7*k-4:

%F k=1: a(n) = a(n-1)

%F k=2: a(n) = a(n-1) +2*a(n-2) -a(n-3) +3*a(n-4) -a(n-5) -7*a(n-6) +a(n-7) +a(n-10)

%F k=3: [order 17]

%F k=4: [order 24]

%F k=5: [order 31]

%F k=6: [order 38]

%F k=7: [order 45]

%F Empirical for row n, apparently polynomial of degree n+3:

%F n=1: a(n) = n^4 + 2*n^3 + 2*n^2 + n

%F n=2: a(n) = n^5 + (3/2)*n^4 + 2*n^3 + (3/2)*n^2

%F n=3: a(n) = n^6 + (16/15)*n^5 + (13/6)*n^4 + (5/3)*n^3 - (1/6)*n^2 + (4/15)*n

%F n=4: [polynomial of degree 7]

%F n=5: [polynomial of degree 8]

%F n=6: [polynomial of degree 9]

%F n=7: [polynomial of degree 10]

%e Some solutions for n=4 k=4

%e ..3....4....1....4....0....4....2....1....1....1....0....0....1....3....4....0

%e ..2....2....2....2....1....0....3....2....4....3....1....0....1....0....0....3

%e ..3....3....0....4....2....2....4....3....3....2....3....3....0....3....4....0

%e ..1....1....0....2....3....1....0....0....2....4....2....2....0....0....0....2

%e ..3....0....4....3....4....4....4....1....4....1....3....1....1....4....2....1

%e ..0....4....4....2....4....0....1....0....4....4....0....3....2....0....3....4

%e ..4....2....2....3....2....3....0....2....3....0....0....4....2....1....0....1

%Y Row 1 is A082986

%K nonn,tabl

%O 1,1

%A _R. H. Hardin_, Nov 20 2014

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