login
Number of length 1+4 0..n arrays with every five consecutive terms having four times some element equal to the sum of the remaining four.
1

%I #8 Nov 09 2018 13:53:01

%S 2,53,164,485,1046,2077,3708,6149,9610,14441,20832,29173,39794,53105,

%T 69476,89417,113318,141769,175260,214361,259642,311813,371424,439225,

%U 515906,602237,698948,806909,926870,1059781,1206492,1367933,1545034

%N Number of length 1+4 0..n arrays with every five consecutive terms having four times some element equal to the sum of the remaining four.

%H R. H. Hardin, <a href="/A249657/b249657.txt">Table of n, a(n) for n = 1..210</a>

%F Empirical: a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) - 2*a(n-4) + 2*a(n-6) - 2*a(n-7) + 3*a(n-8) - 3*a(n-9) + a(n-10).

%F Empirical for n mod 12 = 0: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + 1

%F Empirical for n mod 12 = 1: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n - (53/12)

%F Empirical for n mod 12 = 2: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + (23/3)

%F Empirical for n mod 12 = 3: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n - (31/4)

%F Empirical for n mod 12 = 4: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + (43/3)

%F Empirical for n mod 12 = 5: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n - (133/12)

%F Empirical for n mod 12 = 6: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + 1

%F Empirical for n mod 12 = 7: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + (67/12)

%F Empirical for n mod 12 = 8: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + (23/3)

%F Empirical for n mod 12 = 9: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n - (71/4)

%F Empirical for n mod 12 = 10: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + (43/3)

%F Empirical for n mod 12 = 11: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n - (13/12)

%F Empirical g.f.: x*(2 + 47*x + 11*x^2 + 148*x^3 - 19*x^4 + 172*x^5 - 31*x^6 + 32*x^7 - 3*x^8 + x^9) / ((1 - x)^5*(1 + x)*(1 + x^2)*(1 + x + x^2)). - _Colin Barker_, Nov 09 2018

%e Some solutions for n=6:

%e ..3....0....5....2....2....2....3....4....2....5....4....0....6....6....0....3

%e ..5....0....4....1....3....3....0....6....2....1....3....3....4....2....0....3

%e ..3....6....6....3....2....6....6....6....4....6....1....3....3....4....0....3

%e ..4....6....3....2....1....3....2....2....2....3....2....3....6....6....1....0

%e ..0....3....2....2....2....1....4....2....0....0....5....6....1....2....4....6

%Y Row 1 of A249656.

%K nonn

%O 1,1

%A _R. H. Hardin_, Nov 03 2014