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Number of length 1+3 0..n arrays with every four consecutive terms having the sum of some three elements equal to three times the fourth.
1

%I #8 Nov 08 2018 19:32:53

%S 2,15,52,101,198,331,512,753,1066,1439,1908,2461,3110,3867,4744,5729,

%T 6858,8119,9524,11085,12814,14699,16776,19033,21482,24135,27004,30077,

%U 33390,36931,40712,44745,49042,53591,58428,63541,68942,74643,80656,86969

%N Number of length 1+3 0..n arrays with every four consecutive terms having the sum of some three elements equal to three times the fourth.

%H R. H. Hardin, <a href="/A248538/b248538.txt">Table of n, a(n) for n = 1..210</a>

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

%F Empirical for n mod 6 = 0: a(n) = (4/3)*n^3 + n^2 + n + 1.

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

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

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

%F Empirical for n mod 6 = 4: a(n) = (4/3)*n^3 + n^2 + n - (13/3).

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

%F Empirical g.f.: x*(2 + 11*x + 22*x^2 - x^3 + 13*x^4 + 2*x^5 - x^6) / ((1 - x)^4*(1 + x)*(1 + x + x^2)). - _Colin Barker_, Nov 08 2018

%e Some solutions for n=6:

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

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

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

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

%Y Row 1 of A248537.

%K nonn

%O 1,1

%A _R. H. Hardin_, Oct 08 2014