%I #7 May 30 2019 13:05:28
%S 2,3,4,4,9,8,5,16,27,15,6,25,64,78,28,7,36,125,250,222,51,8,49,216,
%T 615,964,622,92,9,64,343,1281,2995,3674,1722,164,10,81,512,2380,7536,
%U 14455,13868,4719,290,11,100,729,4068,16408,44021,69235,51917,12821,509,12,121
%N T(n,k)=Number of length-n 0..k arrays with no repeated value greater than the previous repeated value.
%C Table starts
%C ...2.....3......4.......5........6.........7.........8..........9.........10
%C ...4.....9.....16......25.......36........49........64.........81........100
%C ...8....27.....64.....125......216.......343.......512........729.......1000
%C ..15....78....250.....615.....1281......2380......4068.......6525.......9955
%C ..28...222....964....2995.....7536.....16408.....32152......58149......98740
%C ..51...622...3674...14455....44021....112476....252932.....516189.....976135
%C ..92..1722..13868...69235...255576....767172...1981512....4566213....9621220
%C .164..4719..51917..329430..1475871...5209554..15465934...40265487...94574110
%C .290.12821.192980.1558430..8482276..35236110.120310016..354051015..927338710
%C .509.34575.712868.7334806.48543777.237479970.933059856.3105016479.9072298237
%C From _Robert Israel_, May 30 2019: (Start)
%C For each of the A000110 partitions pi of the set {1,...,n}, let A_pi(n,k) be the number of length-n 0..k arrays v, such that v(i)=v(j) if and only if i and j are in the same part, and with no repeated value greater than the previous repeated value. There are restrictions on the values in the parts: if two parts a and b each have cardinality >= 2 and a_2 < b_2 (where the parts are indexed in increasing order), then v(b_i) < v(a_i). Thus if there are m partitions with cardinality >= 2, the values on those m parts are decreasing (listing these parts in order of their second entries). So for a partition with j parts of which m have cardinality >= 2, we have A_pi(n,k) = (k+1)*k*...*(k+2-j)/m!, which is a polynomial in k of degree j. The partition of largest cardinality is the partition into singletons, which has m=0. The result is that for each n, T(n,k) is a monic polynomial of degree n. To verify the "empirical" formula for a row, only n terms in that row need to be computed. (End)
%H R. H. Hardin, <a href="/A269435/b269435.txt">Table of n, a(n) for n = 1..9999</a>
%F Empirical for column k:
%F k=1: a(n) = 2*a(n-1) +a(n-2) -2*a(n-3) -a(n-4)
%F k=2: a(n) = 6*a(n-1) -9*a(n-2) -4*a(n-3) +9*a(n-4) +6*a(n-5) +a(n-6)
%F k=3: [order 8]
%F k=4: [order 10]
%F k=5: [order 12]
%F k=6: [order 14]
%F k=7: [order 16]
%F Empirical for row n:
%F n=1: a(n) = n + 1
%F n=2: a(n) = n^2 + 2*n + 1
%F n=3: a(n) = n^3 + 3*n^2 + 3*n + 1
%F n=4: a(n) = n^4 + 4*n^3 + (11/2)*n^2 + (7/2)*n + 1
%F n=5: a(n) = n^5 + 5*n^4 + (17/2)*n^3 + 8*n^2 + (9/2)*n + 1
%F n=6: a(n) = n^6 + 6*n^5 + 12*n^4 + (44/3)*n^3 + (23/2)*n^2 + (29/6)*n + 1
%F n=7: a(n) = n^7 + 7*n^6 + 16*n^5 + (71/3)*n^4 + (139/6)*n^3 + (43/3)*n^2 + (35/6)*n + 1
%e Some solutions for n=6 k=4
%e ..1. .2. .3. .3. .2. .2. .2. .4. .3. .2. .3. .4. .3. .2. .1. .2
%e ..4. .3. .3. .4. .0. .0. .4. .3. .1. .0. .2. .3. .4. .0. .3. .4
%e ..3. .0. .1. .3. .3. .0. .3. .4. .3. .4. .3. .3. .3. .3. .4. .0
%e ..3. .4. .3. .2. .1. .0. .3. .2. .1. .3. .4. .3. .2. .0. .0. .0
%e ..3. .1. .4. .2. .2. .2. .2. .4. .3. .0. .3. .3. .3. .4. .4. .2
%e ..2. .4. .3. .1. .3. .3. .0. .4. .2. .1. .2. .0. .0. .3. .4. .3
%Y Column 1 is A029907(n+1).
%Y Column 2 is A268013.
%Y Column 3 is A267975.
%Y Diagonal is A268104.
%Y Row 1 is A000027(n+1).
%Y Row 2 is A000290(n+1).
%Y Row 3 is A000578(n+1).
%K nonn,tabl
%O 1,1
%A _R. H. Hardin_, Feb 26 2016