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T(n,k)=Number of length n+2 0..k arrays with the medians of every three consecutive terms nondecreasing
13

%I #4 Nov 13 2014 10:14:56

%S 8,27,14,64,67,24,125,204,162,41,216,485,632,391,68,343,986,1827,1959,

%T 900,111,512,1799,4368,6902,5696,2026,180,729,3032,9156,19446,24125,

%U 16104,4530,289,1000,4809,17424,46914,79200,81664,45232,9975,460,1331

%N T(n,k)=Number of length n+2 0..k arrays with the medians of every three consecutive terms nondecreasing

%C Table starts

%C ...8....27.....64.....125......216.......343........512........729........1000

%C ..14....67....204.....485......986......1799.......3032.......4809........7270

%C ..24...162....632....1827.....4368......9156......17424......30789.......51304

%C ..41...391...1959....6902....19446.....46914.....100962.....199023......365959

%C ..68...900...5696...24125....79200....217856.....526032....1149057.....2318140

%C .111..2026..16104...81664...311498....974944....2637228....6376143....14100493

%C .180..4530..45232..274901..1219944...4350588...13201680...35373129....85849852

%C .289..9975.124249..899306..4617079..18667931...63266403..187131076...496682670

%C .460.21694.335328.2878124.17036428..77880418..294117016..958537837..2777976392

%C .728.46871.897523.9128858.62297886.322089271.1356124591.4872648817.15429444696

%H R. H. Hardin, <a href="/A250140/b250140.txt">Table of n, a(n) for n = 1..9999</a>

%F Empirical for column k, apparently a recurrence of order 7*k-1:

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

%F k=2: [order 13]

%F k=3: [order 20]

%F k=4: [order 27]

%F k=5: [order 34]

%F k=6: [order 41]

%F k=7: [order 48]

%F Empirical for row n, apparently a polynomial of degree n+2:

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

%F n=2: a(n) = (2/3)*n^4 + (10/3)*n^3 + (16/3)*n^2 + (11/3)*n + 1

%F n=3: a(n) = (13/30)*n^5 + 3*n^4 + (22/3)*n^3 + 8*n^2 + (127/30)*n + 1

%F n=4: [polynomial of degree 6]

%F n=5: [polynomial of degree 7]

%F n=6: [polynomial of degree 8]

%F n=7: [polynomial of degree 9]

%e Some solutions for n=5 k=4

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

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

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

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

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

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

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

%Y Column 1 is A164406(n+2) for n>1

%Y Row 1 is A000578(n+1)

%K nonn,tabl

%O 1,1

%A _R. H. Hardin_, Nov 13 2014