A275176 a(n) = (3 * a(n-3) + a(n-1) * a(n-5)) / a(n-6), a(0) = a(1) = ... = a(5) = 1.

1, 1, 1, 1, 1, 1, 4, 7, 10, 22, 43, 202, 370, 547, 1264, 2521, 11881, 21781, 32221, 74521, 148681, 700744, 1284667, 1900450, 4395442, 8769643, 41331982, 75773530, 112094287, 259256524, 517260241, 2437886161, 4469353561, 6611662441, 15291739441, 30509584561

Offset

0,7

Comments

Inspired by A048736.

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,60,0,0,0,0,-60,0,0,0,0,1).

Formula

G.f.: (1 +x +x^2 +x^3 +x^4 -59*x^5 -56*x^6 -53*x^7 -50*x^8 -38*x^9 +43*x^10 +22*x^11 +10*x^12 +7*x^13 +4*x^14) / ((1 -x)*(1 +x +x^2 +x^3 +x^4)*(1 -59*x^5 +x^10)). - Colin Barker, Jul 19 2016

a(n) = 60*a(n-5) - 60*a(n-10) + a(n-15).

Mathematica

RecurrenceTable[{a[n] == (3 a[n - 3] + a[n - 1] a[n - 5])/a[n - 6], a[1] == 1, a[2] == 1, a[3] == 1, a[4] == 1, a[5] == 1, a[6] == 1}, a, {n, 36}] (* Michael De Vlieger, Jul 19 2016 *)

Prog

(Ruby)
def A(k, l, n)
  a = Array.new(k * 2, 1)
  ary = [1]
  while ary.size < n + 1
    break if (a[1] * a[-1] + a[k] * l) % a[0] > 0
    a = *a[1..-1], (a[1] * a[-1] + a[k] * l) / a[0]
    ary << a[0]
  end
  ary
end
def A275176(n)
  A(3, 3, n)
end
(PARI) Vec((1 +x +x^2 +x^3 +x^4 -59*x^5 -56*x^6 -53*x^7 -50*x^8 -38*x^9 +43*x^10 +22*x^11 +10*x^12 +7*x^13 +4*x^14) / ((1 -x)*(1 +x +x^2 +x^3 +x^4)*(1 -59*x^5 +x^10)) + O(x^50)) \\ Colin Barker, Jul 19 2016

Crossrefs

Cf. A048736, A275173, A275175.

Sequence in context: A161863 A102649 A084386 * A024726 A024948 A296309

Adjacent sequences: A275173 A275174 A275175 * A275177 A275178 A275179

Keyword

nonn,easy

Author

Seiichi Manyama, Jul 19 2016

Status

approved