A221463 - OEIS

%I #6 Jan 21 2013 12:55:55

%S 0,0,1,0,2,1,0,3,4,2,0,4,9,12,3,0,5,16,36,32,5,0,6,25,80,135,88,8,0,7,

%T 36,150,384,513,240,13,0,8,49,252,875,1856,1944,656,21,0,9,64,392,

%U 1728,5125,8960,7371,1792,34,0,10,81,576,3087,11880,30000,43264,27945,4896,55,0,11

%N T(n,k)=Number of 0..k arrays of length n with each element unequal to at least one neighbor, starting with 0

%C Table starts

%C ..0.....0.......0........0.........0..........0..........0...........0

%C ..1.....2.......3........4.........5..........6..........7...........8

%C ..1.....4.......9.......16........25.........36.........49..........64

%C ..2....12......36.......80.......150........252........392.........576

%C ..3....32.....135......384.......875.......1728.......3087........5120

%C ..5....88.....513.....1856......5125......11880......24353.......45568

%C ..8...240....1944.....8960.....30000......81648.....192080......405504

%C .13...656....7371....43264....175625.....561168....1515031.....3608576

%C .21..1792...27945...208896...1028125....3856896...11949777....32112640

%C .34..4896..105948..1008640...6018750...26508384...94253656...285769728

%C .55.13376..401679..4870144..35234375..182191680..743424031..2543058944

%C .89.36544.1522881.23515136.206265625.1252200384.5863743809.22630629376

%H R. H. Hardin, <a href="/A221463/b221463.txt">Table of n, a(n) for n = 1..10000</a>

%F Recursion for column k:

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

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

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

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

%F k=5: a(n) = 5*a(n-1) +5*a(n-2)

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

%F k=7: a(n) = 7*a(n-1) +7*a(n-2)

%F Empirical for row n:

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

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

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

%F n=5: a(k) = 1*k^4 + 2*k^3

%F n=6: a(k) = 1*k^5 + 3*k^4 + 1*k^3

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

%F n=8: a(k) = 1*k^7 + 5*k^6 + 6*k^5 + 1*k^4

%F n=9: a(k) = 1*k^8 + 6*k^7 + 10*k^6 + 4*k^5

%F n=10: a(k) = 1*k^9 + 7*k^8 + 15*k^7 + 10*k^6 + 1*k^5

%F n=11: a(k) = 1*k^10 + 8*k^9 + 21*k^8 + 20*k^7 + 5*k^6

%F n=12: a(k) = 1*k^11 + 9*k^10 + 28*k^9 + 35*k^8 + 15*k^7 + 1*k^6

%F n=13: a(k) = 1*k^12 + 10*k^11 + 36*k^10 + 56*k^9 + 35*k^8 + 6*k^7

%F n=14: a(k) = 1*k^13 + 11*k^12 + 45*k^11 + 84*k^10 + 70*k^9 + 21*k^8 + 1*k^7

%F n=15: a(k) = 1*k^14 + 12*k^13 + 55*k^12 + 120*k^11 + 126*k^10 + 56*k^9 + 7*k^8

%F Apparently then T(n,k) = sum { binomial(n-2-i,i)*k^(n-1-i) , 0<=2*i<=n-2 }.

%F The formula reduces to T(n,k) = [4*k^(n-1)*(1+G)^(2*n-2) +4^n] /[2^(n+1) *G *(1+G)^(n-1)] for even n and to T(n,k) = [4*k^(n-1) *(1+G)^(2*n-2) -4^n] /[2^(n+1) *G *(1+G)^(n-1)] for odd n, where G=sqrt(1+4/k). - _R. J. Mathar_, Jan 21 2013

%e Some solutions for n=6 k=4

%e ..0....0....0....0....0....0....0....0....0....0....0....0....0....0....0....0

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

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

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

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

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

%Y Column 1 is A000045(n-1)

%Y Column 2 is A028860(n+1)

%Y Column 3 is A106435(n-1)

%Y Column 4 is A094013

%Y Column 5 is A106565(n-1)

%Y Row 2 is A000027

%Y Row 3 is A000290

%Y Row 4 is A011379

%K nonn,tabl

%O 1,5

%A _R. H. Hardin_, general recursion proved by _Robert Israel_ in the Sequence Fans Mailing List, Jan 17 2013