A249320 Number of length 1+6 0..n arrays with no seven consecutive terms having six times any element equal to the sum of the remaining six.

126, 1792, 14336, 68712, 249088, 739284, 1898582, 4361056, 9176664, 17982930, 33230442, 58450770, 98581896, 160353634, 252737912, 387464994, 579621658, 848320494, 1217462624, 1716582952, 2381796186, 3256837500, 4394213292

Offset

1,1

Comments

Row 1 of A249319.

Links

R. H. Hardin, Table of n, a(n) for n = 1..52

Formula

Empirical: a(n) = 6*a(n-1) -16*a(n-2) +26*a(n-3) -30*a(n-4) +27*a(n-5) -21*a(n-6) +16*a(n-7) -11*a(n-8) +4*a(n-9) +4*a(n-10) -11*a(n-11) +16*a(n-12) -21*a(n-13) +27*a(n-14) -30*a(n-15) +26*a(n-16) -16*a(n-17) +6*a(n-18) -a(n-19).

Also a polynomial of degree 7 plus a degree 0 quasipolynomial with period 60; the first 12 are:

Empirical for n mod 60 = 0: a(n) = n^7 + (35/6)*n^6 + (91/5)*n^5 + (119/4)*n^4 + (91/3)*n^3 + (35/2)*n^2 + 6*n.

Empirical for n mod 60 = 1: a(n) = n^7 + (35/6)*n^6 + (91/5)*n^5 + (119/4)*n^4 + (91/3)*n^3 + (35/2)*n^2 + 6*n + (1043/60).

Empirical for n mod 60 = 2: a(n) = n^7 + (35/6)*n^6 + (91/5)*n^5 + (119/4)*n^4 + (91/3)*n^3 + (35/2)*n^2 + 6*n - (462/5).

Empirical for n mod 60 = 3: a(n) = n^7 + (35/6)*n^6 + (91/5)*n^5 + (119/4)*n^4 + (91/3)*n^3 + (35/2)*n^2 + 6*n + (1393/20).

Empirical for n mod 60 = 4: a(n) = n^7 + (35/6)*n^6 + (91/5)*n^5 + (119/4)*n^4 + (91/3)*n^3 + (35/2)*n^2 + 6*n - (952/15).

Empirical for n mod 60 = 5: a(n) = n^7 + (35/6)*n^6 + (91/5)*n^5 + (119/4)*n^4 + (91/3)*n^3 + (35/2)*n^2 + 6*n + (357/4).

Empirical for n mod 60 = 6: a(n) = n^7 + (35/6)*n^6 + (91/5)*n^5 + (119/4)*n^4 + (91/3)*n^3 + (35/2)*n^2 + 6*n - (546/5).

Empirical for n mod 60 = 7: a(n) = n^7 + (35/6)*n^6 + (91/5)*n^5 + (119/4)*n^4 + (91/3)*n^3 + (35/2)*n^2 + 6*n + (7931/60).

Empirical for n mod 60 = 8: a(n) = n^7 + (35/6)*n^6 + (91/5)*n^5 + (119/4)*n^4 + (91/3)*n^3 + (35/2)*n^2 + 6*n - (1008/5).

Empirical for n mod 60 = 9: a(n) = n^7 + (35/6)*n^6 + (91/5)*n^5 + (119/4)*n^4 + (91/3)*n^3 + (35/2)*n^2 + 6*n + (3129/20).

Empirical for n mod 60 = 10: a(n) = n^7 + (35/6)*n^6 + (91/5)*n^5 + (119/4)*n^4 + (91/3)*n^3 + (35/2)*n^2 + 6*n - (140/3).

Empirical for n mod 60 = 11: a(n) = n^7 + (35/6)*n^6 + (91/5)*n^5 + (119/4)*n^4 + (91/3)*n^3 + (35/2)*n^2 + 6*n - (1799/20).

Example

Some solutions for n=4
..3....4....1....3....3....1....4....0....1....0....3....2....4....1....4....4
..0....4....2....2....0....2....0....1....0....3....1....4....0....1....2....3
..1....4....4....1....4....1....4....3....3....0....3....2....0....0....2....4
..3....1....3....2....2....2....3....4....2....4....2....1....1....0....0....1
..3....3....1....4....2....4....3....1....1....2....0....4....3....0....0....4
..1....3....0....4....1....1....2....4....4....0....3....0....2....2....2....1
..0....3....4....2....3....0....2....1....2....2....3....0....2....0....1....2

Crossrefs

Cf. A249319.

Sequence in context: A249319 A249198 A249213 * A267628 A304559 A189345

Adjacent sequences: A249317 A249318 A249319 * A249321 A249322 A249323

Keyword

nonn

Author

R. H. Hardin, Oct 25 2014

Status

approved