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.
62, 606, 3492, 13580, 40950, 104562, 235196, 480912, 911490, 1625210, 2754672, 4474896, 7011302, 10648350, 15739200, 22716332, 32101806, 44520162, 60709580, 81535980, 108006522, 141284426, 182704032, 233787120, 296259530, 372068862
Offset: 1
Keywords
Examples
Some solutions for n=6 ..5....0....2....3....3....4....5....1....4....5....0....0....2....3....4....0 ..2....4....5....1....1....6....1....5....4....4....2....4....5....5....1....2 ..0....5....1....0....2....3....1....3....6....2....1....4....4....3....6....3 ..1....3....0....4....5....2....1....2....2....6....1....4....1....0....2....1 ..0....3....2....0....2....4....3....5....1....6....6....0....3....0....4....5 ..0....1....3....6....1....3....6....1....6....2....6....0....4....3....2....6
Links
- R. H. Hardin, Table of n, a(n) for n = 1..101
Formula
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)
Also a degree 6 polynomial plus a degree 0 quasipolynomial with period 60, the first 12 being:
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
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
Comments