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A238188
a(n) = 4*a(n-4) + 6*a(n-8) + 4*a(n-12) + a(n-16) for n>15, with the sixteen initial values as shown.
1
0, 0, 1, 1, 2, 2, 2, 3, 9, 11, 13, 15, 48, 57, 68, 81, 254, 302, 359, 427, 1342, 1596, 1898, 2257, 7093, 8435, 10031, 11929, 37488, 44581, 53016, 63047, 198132, 235620, 280201, 333217, 1047170, 1245302, 1480922, 1761123, 5534517, 6581687, 7826989, 9307911, 29251104, 34785621, 41367308
OFFSET
0,5
COMMENTS
These four sequences:
b(4n+4) = b(4n) + b(4n+1) + b(4n+2) + b(4n+3),
b(4n+5) = 2*b(4n) + b(4n+1) + b(4n+2) + b(4n+3),
b(4n+6) = 2*b(4n) + 2*b(4n+1) + b(4n+2) + b(4n+3),
b(4n+7) = 2*b(4n) + 2*b(4n+1) + 2*b(4n+2) + b(4n+3),
give the polynomial: x^4-4*x^3-6*x^2-4*x-1 with root 1 + 2^(1/4) + 2^(2/4) + 2^(3/4). More generally, see link the roots of the equation of the fourth degree.
Equation: 8*x^4-t^4-2*(-z^4+4*t*y*z^2-4*t^2*x*z-2*t^2*y^2)-4*(x^2*(2*z^2+4*t*y)-4*x*y^2*z+y^4) = (-1)^(ceiling(n/2)-floor(n/2)), if x = a(4n), y = a(4n+1), z = a(4n+2), t = a(4n+3).
LINKS
Alexander Samokrutov, Table of n, a(n) for n = 0..91
Index entries for linear recurrences with constant coefficients, signature (0,0,0,4,0,0,0,6,0,0,0,4,0,0,0,1).
FORMULA
G.f.: x^2*(x^2+x+1)*(x^3-x^2+1)*(x^8-x^6+2*x^4-2*x^2-1) / (x^16+4*x^12+6*x^8+4*x^4-1). - Colin Barker, May 02 2015
MATHEMATICA
LinearRecurrence[{0, 0, 0, 4, 0, 0, 0, 6, 0, 0, 0, 4, 0, 0, 0, 1}, {0, 0, 1, 1, 2, 2, 2, 3, 9, 11, 13, 15, 48, 57, 68, 81}, 60] (* Vincenzo Librandi, May 15 2015 *)
PROG
(PARI) concat([0, 0], Vec(x^2*(x^2+x+1)*(x^3-x^2+1)*(x^8-x^6+2*x^4-2*x^2-1) / (x^16+4*x^12+6*x^8+4*x^4-1) + O(x^100))) \\ Colin Barker, May 02 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved