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%I #6 Dec 12 2014 20:58:44
%S 22,183,988,3301,8370,17923,33520,55665,87310,129991,185340,256861,
%T 346258,456195,590920,751873,942030,1167247,1428940,1730397,2078794,
%U 2475475,2924280,3433177,4003774,4640463,5351908,6140413,7010538,7972267
%N Number of length 3+5 0..n arrays with some three disjoint pairs in each consecutive six terms having the same sum
%C Row 3 of A248489
%H R. H. Hardin, <a href="/A248492/b248492.txt">Table of n, a(n) for n = 1..210</a>
%F Empirical: a(n) = -a(n-2) +a(n-3) +2*a(n-5) +a(n-6) +2*a(n-7) +a(n-8) -a(n-10) -3*a(n-11) -2*a(n-12) -4*a(n-13) -a(n-14) -2*a(n-15) +2*a(n-16) +a(n-17) +4*a(n-18) +2*a(n-19) +3*a(n-20) +a(n-21) -a(n-23) -2*a(n-24) -a(n-25) -2*a(n-26) -a(n-28) +a(n-29) +a(n-31)
%F Also a quartic polynomial plus a linear quasipolynomial with period 840, the first 12 being:
%F Empirical for n mod 840 = 0: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (1769/35)*n + 1
%F Empirical for n mod 840 = 1: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (1664/35)*n + (133093/630)
%F Empirical for n mod 840 = 2: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (10907/105)*n + (175159/315)
%F Empirical for n mod 840 = 3: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (2084/35)*n + (7001/14)
%F Empirical for n mod 840 = 4: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (1769/35)*n + (73067/315)
%F Empirical for n mod 840 = 5: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (10592/105)*n + (2311/18)
%F Empirical for n mod 840 = 6: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (1769/35)*n - (4969/35)
%F Empirical for n mod 840 = 7: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (2084/35)*n + (30739/90)
%F Empirical for n mod 840 = 8: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (10907/105)*n + (27191/63)
%F Empirical for n mod 840 = 9: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (1664/35)*n + (10141/70)
%F Empirical for n mod 840 = 10: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (1769/35)*n + (20035/63)
%F Empirical for n mod 840 = 11: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (11852/105)*n + (367733/630)
%e Some solutions for n=6
%e ..5....6....4....4....4....2....4....6....4....0....4....5....3....3....6....3
%e ..2....5....3....0....0....2....6....5....2....2....3....4....3....5....1....4
%e ..2....4....6....0....1....3....5....4....4....1....2....4....4....4....2....5
%e ..1....5....2....2....2....4....4....1....0....0....3....5....5....2....5....3
%e ..5....6....5....4....3....4....3....2....6....1....2....3....1....1....4....4
%e ..6....4....4....2....2....3....2....3....2....2....1....3....2....3....3....2
%e ..5....3....1....4....1....5....1....0....4....3....4....2....3....3....0....3
%e ..2....2....0....0....3....2....0....5....2....2....3....4....6....5....1....4
%K nonn
%O 1,1
%A _R. H. Hardin_, Oct 07 2014
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