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%I #6 Dec 12 2014 20:58:15
%S 16,72,460,1512,4272,9684,20236,37868,67140,111104,177024,269892,
%T 400040,574140,806808,1107132,1493952,1978984,2586340,3330852,4242444,
%U 5338788,6656352,8217136,10064140,12222784,14744480,17659176,21025952,24879392
%N Number of length 3+2 0..n arrays with no three consecutive terms having the sum of any two elements equal to twice the third
%C Row 3 of A248461
%H R. H. Hardin, <a href="/A248464/b248464.txt">Table of n, a(n) for n = 1..210</a>
%F Empirical: a(n) = a(n-1) +a(n-3) -2*a(n-7) +a(n-8) -2*a(n-9) +a(n-10) -a(n-11) +2*a(n-12) +2*a(n-15) -a(n-16) +a(n-17) -2*a(n-18) +a(n-19) -2*a(n-20) +a(n-24) +a(n-26) -a(n-27)
%F Also a polynomial of degree 5 plus a quadratic quasipolynomial with period 840, the first 12 being:
%F Empirical for n mod 840 = 0: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (51/5)*n
%F Empirical for n mod 840 = 1: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (66/5)*n - (33/10)
%F Empirical for n mod 840 = 2: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (113/15)*n - (152/15)
%F Empirical for n mod 840 = 3: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (26/5)*n - (1/2)
%F Empirical for n mod 840 = 4: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (51/5)*n - (12/5)
%F Empirical for n mod 840 = 5: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (158/15)*n + (1/6)
%F Empirical for n mod 840 = 6: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (51/5)*n - (24/5)
%F Empirical for n mod 840 = 7: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (26/5)*n + (67/10)
%F Empirical for n mod 840 = 8: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (113/15)*n - (4/3)
%F Empirical for n mod 840 = 9: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (66/5)*n - (9/10)
%F Empirical for n mod 840 = 10: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (51/5)*n - 8
%F Empirical for n mod 840 = 11: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (38/15)*n + (41/30)
%e Some solutions for n=6
%e ..2....2....3....1....4....2....2....1....2....4....0....1....3....3....6....5
%e ..3....5....2....3....1....6....6....4....0....4....2....6....6....0....0....4
%e ..5....6....0....1....4....1....2....4....0....1....5....5....6....5....2....4
%e ..2....0....5....6....4....6....5....0....3....5....3....0....2....5....2....0
%e ..4....5....3....2....1....4....2....6....5....5....6....5....2....6....3....5
%K nonn
%O 1,1
%A _R. H. Hardin_, Oct 06 2014
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