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T(n,k)=Number of length-n 0..k arrays with no repeated value equal to the previous repeated value.
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%I #10 Nov 02 2024 09:13:26

%S 2,3,4,4,9,6,5,16,24,10,6,25,60,66,14,7,36,120,228,174,22,8,49,210,

%T 580,852,462,30,9,64,336,1230,2780,3180,1206,46,10,81,504,2310,7170,

%U 13300,11796,3150,62,11,100,720,3976,15834,41730,63420,43644,8166,94,12,121

%N T(n,k)=Number of length-n 0..k arrays with no repeated value equal to 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 ..6....24.....60.....120......210.......336.......504........720........990

%C .10....66....228.....580.....1230......2310......3976.......6408.......9810

%C .14...174....852....2780.....7170.....15834.....31304......56952......97110

%C .22...462...3180...13300....41730....108402....246232.....505800.....960750

%C .30..1206..11796...63420...242370....741090...1934856....4488696....9499590

%C .46..3150..43644..301780..1405530...5060706..15190840...39808584...93880710

%C .62..8166.160980.1433180..8139570..34523202.119174216..352838520..927352710

%C .94.21150.592572.6795700.47082330.235304034.934305400.3125681352.9156504150

%C The conjectures regarding the recursions for column k are correct (see links) - _Sela Fried_, Oct 29 2024.

%H R. H. Hardin, <a href="/A269467/b269467.txt">Table of n, a(n) for n = 1..9999</a>

%H Sela Fried, <a href="https://arxiv.org/abs/2410.07237">Proofs of some Conjectures from the OEIS</a>, arXiv:2410.07237 [math.NT], 2024.

%F Empirical for column k:

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

%F k=2: a(n) = 3*a(n-1) +2*a(n-2) -8*a(n-3)

%F k=3: a(n) = 5*a(n-1) -18*a(n-3)

%F k=4: a(n) = 7*a(n-1) -4*a(n-2) -32*a(n-3)

%F k=5: a(n) = 9*a(n-1) -10*a(n-2) -50*a(n-3)

%F k=6: a(n) = 11*a(n-1) -18*a(n-2) -72*a(n-3)

%F k=7: a(n) = 13*a(n-1) -28*a(n-2) -98*a(n-3)

%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 + 2*n

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

%F n=5: a(n) = n^5 + 5*n^4 + 7*n^3 + 2*n^2 - n

%F n=6: a(n) = n^6 + 6*n^5 + 11*n^4 + 4*n^3 - n^2 + n

%F n=7: a(n) = n^7 + 7*n^6 + 16*n^5 + 8*n^4 - n^3 - n

%e Some solutions for n=6 k=4

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

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

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

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

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

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

%Y Column 1 is A027383.

%Y Row 1 is A000027(n+1).

%Y Row 2 is A000290(n+1).

%Y Row 3 is A007531(n+2).

%K nonn,tabl

%O 1,1

%A _R. H. Hardin_, Feb 27 2016