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A249531 Number of length 1+5 0..n arrays with no six consecutive terms having five times any element equal to the sum of the remaining five.

%I A249531 #8 Jul 23 2025 12:04:07
%S A249531 62,606,3492,13580,40950,104562,235196,480912,911490,1625210,2754672,
%T A249531 4474896,7011302,10648350,15739200,22716332,32101806,44520162,
%U A249531 60709580,81535980,108006522,141284426,182704032,233787120,296259530,372068862
%N A249531 Number of length 1+5 0..n arrays with no six consecutive terms having five times any element equal to the sum of the remaining five.
%C A249531 Row 1 of A249530
%H A249531 R. H. Hardin, <a href="/A249531/b249531.txt">Table of n, a(n) for n = 1..101</a>
%F A249531 Empirical: a(n) = 4*a(n-1) -6*a(n-2) +5*a(n-3) -4*a(n-4) +3*a(n-5) -2*a(n-6) +2*a(n-7) -2*a(n-9) +2*a(n-10) -3*a(n-11) +4*a(n-12) -5*a(n-13) +6*a(n-14) -4*a(n-15) +a(n-16)
%F A249531 Also a degree 6 polynomial plus a degree 0 quasipolynomial with period 60, the first 12 being:
%F A249531 Empirical for n mod 60 = 0: a(n) = n^6 + (24/5)*n^5 + (51/4)*n^4 + (49/3)*n^3 + 13*n^2 + 5*n
%F A249531 Empirical for n mod 60 = 1: a(n) = n^6 + (24/5)*n^5 + (51/4)*n^4 + (49/3)*n^3 + 13*n^2 + 5*n + (547/60)
%F A249531 Empirical for n mod 60 = 2: a(n) = n^6 + (24/5)*n^5 + (51/4)*n^4 + (49/3)*n^3 + 13*n^2 + 5*n - (124/15)
%F A249531 Empirical for n mod 60 = 3: a(n) = n^6 + (24/5)*n^5 + (51/4)*n^4 + (49/3)*n^3 + 13*n^2 + 5*n - (183/20)
%F A249531 Empirical for n mod 60 = 4: a(n) = n^6 + (24/5)*n^5 + (51/4)*n^4 + (49/3)*n^3 + 13*n^2 + 5*n + (472/15)
%F A249531 Empirical for n mod 60 = 5: a(n) = n^6 + (24/5)*n^5 + (51/4)*n^4 + (49/3)*n^3 + 13*n^2 + 5*n - (425/12)
%F A249531 Empirical for n mod 60 = 6: a(n) = n^6 + (24/5)*n^5 + (51/4)*n^4 + (49/3)*n^3 + 13*n^2 + 5*n + (156/5)
%F A249531 Empirical for n mod 60 = 7: a(n) = n^6 + (24/5)*n^5 + (51/4)*n^4 + (49/3)*n^3 + 13*n^2 + 5*n - (821/60)
%F A249531 Empirical for n mod 60 = 8: a(n) = n^6 + (24/5)*n^5 + (51/4)*n^4 + (49/3)*n^3 + 13*n^2 + 5*n + (344/15)
%F A249531 Empirical for n mod 60 = 9: a(n) = n^6 + (24/5)*n^5 + (51/4)*n^4 + (49/3)*n^3 + 13*n^2 + 5*n - (879/20)
%F A249531 Empirical for n mod 60 = 10: a(n) = n^6 + (24/5)*n^5 + (51/4)*n^4 + (49/3)*n^3 + 13*n^2 + 5*n + (80/3)
%F A249531 Empirical for n mod 60 = 11: a(n) = n^6 + (24/5)*n^5 + (51/4)*n^4 + (49/3)*n^3 + 13*n^2 + 5*n + (1547/60)
%e A249531 Some solutions for n=6
%e A249531 ..5....0....2....3....3....4....5....1....4....5....0....0....2....3....4....0
%e A249531 ..2....4....5....1....1....6....1....5....4....4....2....4....5....5....1....2
%e A249531 ..0....5....1....0....2....3....1....3....6....2....1....4....4....3....6....3
%e A249531 ..1....3....0....4....5....2....1....2....2....6....1....4....1....0....2....1
%e A249531 ..0....3....2....0....2....4....3....5....1....6....6....0....3....0....4....5
%e A249531 ..0....1....3....6....1....3....6....1....6....2....6....0....4....3....2....6
%K A249531 nonn
%O A249531 1,1
%A A249531 _R. H. Hardin_, Oct 31 2014