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A247929 Number of length 2+4 0..n arrays with some disjoint pairs in every consecutive five terms having the same sum 1

%I #6 Dec 12 2014 20:57:07

%S 64,529,2760,9569,25512,57769,117256,216937,376656,613721,956736,

%T 1432465,2078080,2925313,4025112,5420273,7170312,9325369,11960584,

%U 15142537,18952080,23466953,28787520,35002753,42229648,50570737,60151800,71100353

%N Number of length 2+4 0..n arrays with some disjoint pairs in every consecutive five terms having the same sum

%C Row 2 of A247927

%H R. H. Hardin, <a href="/A247929/b247929.txt">Table of n, a(n) for n = 1..152</a>

%F Empirical: a(n) = a(n-2) +a(n-3) +a(n-4) -a(n-7) -2*a(n-8) -2*a(n-9) -a(n-10) +2*a(n-12) +2*a(n-13) +2*a(n-14) +2*a(n-15) -a(n-17) -2*a(n-18) -2*a(n-19) -a(n-20) +a(n-23) +a(n-24) +a(n-25) -a(n-27)

%F Also a polynomial of degree 5 plus a linear quasipolynomial with period 420; the first 12 are :

%F Empirical for n mod 420 = 0: a(n) = 2*n^5 + 75*n^4 - (1460/3)*n^3 + (58568/35)*n^2 - (60384/35)*n + 1

%F Empirical for n mod 420 = 1: a(n) = 2*n^5 + 75*n^4 - (1460/3)*n^3 + (58358/35)*n^2 - (50934/35)*n + (27463/105)

%F Empirical for n mod 420 = 2: a(n) = 2*n^5 + 75*n^4 - (1460/3)*n^3 + (58568/35)*n^2 - (57024/35)*n - (29047/105)

%F Empirical for n mod 420 = 3: a(n) = 2*n^5 + 75*n^4 - (1460/3)*n^3 + (58358/35)*n^2 - (50934/35)*n - (9111/7)

%F Empirical for n mod 420 = 4: a(n) = 2*n^5 + 75*n^4 - (1460/3)*n^3 + (58568/35)*n^2 - (60384/35)*n - (42551/105)

%F Empirical for n mod 420 = 5: a(n) = 2*n^5 + 75*n^4 - (1460/3)*n^3 + (58358/35)*n^2 - (47574/35)*n - (5003/3)

%F Empirical for n mod 420 = 6: a(n) = 2*n^5 + 75*n^4 - (1460/3)*n^3 + (58568/35)*n^2 - (60384/35)*n + (8651/35)

%F Empirical for n mod 420 = 7: a(n) = 2*n^5 + 75*n^4 - (1460/3)*n^3 + (58358/35)*n^2 - (50934/35)*n - (15311/15)

%F Empirical for n mod 420 = 8: a(n) = 2*n^5 + 75*n^4 - (1460/3)*n^3 + (58568/35)*n^2 - (57024/35)*n - (14435/21)

%F Empirical for n mod 420 = 9: a(n) = 2*n^5 + 75*n^4 - (1460/3)*n^3 + (58358/35)*n^2 - (50934/35)*n - (24387/35)

%F Empirical for n mod 420 = 10: a(n) = 2*n^5 + 75*n^4 - (1460/3)*n^3 + (58568/35)*n^2 - (60384/35)*n + (6365/21)

%F Empirical for n mod 420 = 11: a(n) = 2*n^5 + 75*n^4 - (1460/3)*n^3 + (58358/35)*n^2 - (47574/35)*n - (261217/105)

%e Some solutions for n=6

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

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

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

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

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

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

%K nonn

%O 1,1

%A _R. H. Hardin_, Sep 26 2014

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