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A122504
a(n) = -a(n-6) + 3*a(n-5) + a(n-4) - 7*a(n-3) + a(n-2) + 3*a(n-1).
0
1, 1, 1, 1, 1, 1, 0, -3, -13, -39, -107, -273, -675, -1624, -3847, -8995, -20851, -47995, -109915, -250695, -570024, -1292915, -2926953, -6616051, -14936895, -33690357, -75931283, -171029936, -385046687, -866536007, -1949510615, -4384874471, -9860587191, -22170707871, -49842661456
OFFSET
1,8
COMMENTS
Original name started "Bi_Steinbach heptagon recursion".
FORMULA
O.g.f.: x*(1-x-3*x^2)*(1-x-x^2)/((1-2*x-x^2+x^3)*(1-x-2*x^2+x^3)). - R. J. Mathar, Aug 22 2008
MATHEMATICA
a[0] = 1; a[1] = 1; a[2] = 1; a[3] = 1; a[4] = 1; a[5] = 1; a[n_] := a[n] = -a[n - 6] + 3 a[n - 5] + a[n - 4] - 7 a[n - 3] + a[n - 2] + 3 a[n - 1] Table[a[n], {n, 0, 30}]
M = {{0, 1, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0}, {0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 1}, {-1, 3, 1, -7, 1, 3}} v[1] = {1, 1, 1, 1, 1, 1} v[n_] := v[n] = M.v[n - 1] a = Table[Floor[v[n][[1]]], {n, 1, 50}]
LinearRecurrence[{3, 1, -7, 1, 3, -1}, {1, 1, 1, 1, 1, 1}, 40] (* or *) Rest[ CoefficientList[ Series[x(1-x-3x^2)(1-x-x^2)/((1-2x-x^2+x^3)(1-x-2x^2+x^3)), {x, 0, 40}], x]] (* Harvey P. Dale, Jun 24 2011 *)
CROSSREFS
Sequence in context: A166911 A103657 A320661 * A103277 A166897 A167910
KEYWORD
sign
AUTHOR
STATUS
approved