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Number of length 2+5 0..n arrays with no three disjoint pairs in any consecutive six terms having the same sum
1

%I #4 Oct 06 2014 18:46:47

%S 62,1272,11436,59480,226410,694632,1824272,4257336,9061830,17909120,

%T 33303852,58859112,99630266,162505920,256670520,394127072,590308422,

%U 864758736,1241905340,1751918880,2431676802,3325811912,4487885496

%N Number of length 2+5 0..n arrays with no three disjoint pairs in any consecutive six terms having the same sum

%C Row 2 of A248448

%H R. H. Hardin, <a href="/A248450/b248450.txt">Table of n, a(n) for n = 1..54</a>

%F Empirical: a(n) = 4*a(n-1) -5*a(n-2) +2*a(n-3) -2*a(n-4) +2*a(n-5) +5*a(n-6) -8*a(n-7) +6*a(n-8) -8*a(n-9) +5*a(n-10) +2*a(n-11) -2*a(n-12) +2*a(n-13) -5*a(n-14) +4*a(n-15) -a(n-16)

%F Empirical for n mod 12 = 0: a(n) = n^7 + 7*n^6 + 6*n^5 + (55/2)*n^4 + (415/9)*n^3 - (383/3)*n^2 + 74*n

%F Empirical for n mod 12 = 1: a(n) = n^7 + 7*n^6 + 6*n^5 + (55/2)*n^4 + (415/9)*n^3 - (383/3)*n^2 + 89*n + (235/18)

%F Empirical for n mod 12 = 2: a(n) = n^7 + 7*n^6 + 6*n^5 + (55/2)*n^4 + (415/9)*n^3 - (383/3)*n^2 + (142/3)*n + (1000/9)

%F Empirical for n mod 12 = 3: a(n) = n^7 + 7*n^6 + 6*n^5 + (55/2)*n^4 + (415/9)*n^3 - (383/3)*n^2 + 89*n + (195/2)

%F Empirical for n mod 12 = 4: a(n) = n^7 + 7*n^6 + 6*n^5 + (55/2)*n^4 + (415/9)*n^3 - (383/3)*n^2 + 74*n + (320/9)

%F Empirical for n mod 12 = 5: a(n) = n^7 + 7*n^6 + 6*n^5 + (55/2)*n^4 + (415/9)*n^3 - (383/3)*n^2 + (187/3)*n + (1595/18)

%F Empirical for n mod 12 = 6: a(n) = n^7 + 7*n^6 + 6*n^5 + (55/2)*n^4 + (415/9)*n^3 - (383/3)*n^2 + 74*n

%F Empirical for n mod 12 = 7: a(n) = n^7 + 7*n^6 + 6*n^5 + (55/2)*n^4 + (415/9)*n^3 - (383/3)*n^2 + 89*n + (2395/18)

%F Empirical for n mod 12 = 8: a(n) = n^7 + 7*n^6 + 6*n^5 + (55/2)*n^4 + (415/9)*n^3 - (383/3)*n^2 + (142/3)*n + (1000/9)

%F Empirical for n mod 12 = 9: a(n) = n^7 + 7*n^6 + 6*n^5 + (55/2)*n^4 + (415/9)*n^3 - (383/3)*n^2 + 89*n - (45/2)

%F Empirical for n mod 12 = 10: a(n) = n^7 + 7*n^6 + 6*n^5 + (55/2)*n^4 + (415/9)*n^3 - (383/3)*n^2 + 74*n + (320/9)

%F Empirical for n mod 12 = 11: a(n) = n^7 + 7*n^6 + 6*n^5 + (55/2)*n^4 + (415/9)*n^3 - (383/3)*n^2 + (187/3)*n + (3755/18)

%e Some solutions for n=4

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

%e ..0....1....0....4....4....1....0....1....1....2....2....1....0....2....4....4

%e ..2....4....1....4....0....1....2....1....0....1....2....0....1....0....0....0

%e ..2....0....3....3....3....3....3....3....3....0....3....3....0....3....4....3

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

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

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

%K nonn

%O 1,1

%A _R. H. Hardin_, Oct 06 2014