A276529 a(n) = (a(n-1) * a(n-5) + 1) / a(n-6), a(0) = a(1) = ... = a(5) = 1.

1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 13, 20, 27, 34, 41, 89, 137, 185, 233, 281, 610, 939, 1268, 1597, 1926, 4181, 6436, 8691, 10946, 13201, 28657, 44113, 59569, 75025, 90481, 196418, 302355, 408292, 514229, 620166, 1346269, 2072372, 2798475, 3524578, 4250681, 9227465

Offset: 0

Author

Seiichi Manyama, Nov 16 2016

Keywords

• nonn
• easy

Comments

Thanks to the linear recurrence signature, we see that this is actually five separate linear recurrence sequences, each with signature (7,-1), interwoven together. - Greg Dresden, Oct 16 2021

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..5985
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,7,0,0,0,0,-1).

Crossrefs

Cf. A275173, A102276, A276530.

5th-sections: A049685, A033891, A033889.

Programs

Mathematica

LinearRecurrence[{0,0,0,0,7,0,0,0,0,-1}, {1,1,1,1,1,1,2,3,4,5}, 50] (* G. C. Greubel, Nov 18 2016 *)
RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==a[4]==a[5]==1,a[n]==(a[n-1]a[n-5]+ 1)/a[n-6]},a,{n,50}] (* Harvey P. Dale, Oct 08 2020 *)
Flatten[Table[{LucasL[4 n - 2]/3, Fibonacci[4 n - 1], LucasL[4 n + 2]/3 - Fibonacci[4 n], LucasL[4 n - 2]/3 + Fibonacci[4 n], Fibonacci[4 n + 1]}, {n, 0, 10}]] (* Greg Dresden, Oct 16 2021 *)

PARI

Vec((1 +x +x^2 +x^3 +x^4 -6*x^5 -5*x^6 -4*x^7 -3*x^8 -2*x^9)/(1 -7*x^5 +x^10) + O(x^50)) \\ Colin Barker, Nov 16 2016

Ruby

def A(k, m, n)
  a = Array.new(2 * k, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[-1] * a[1] + a[k] ** m
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276529(n)
  A(3, 0, n)
end

Formula

a(n) + a(n+10) = 7*a(n+5).

a(5-n) = a(n).

G.f.: (1 +x +x^2 +x^3 +x^4 -6*x^5 -5*x^6 -4*x^7 -3*x^8 -2*x^9) / (1 -7*x^5 +x^10). - Colin Barker, Nov 16 2016

From Greg Dresden, Oct 16 2021: (Start)

a(5*n) = L(4*n-2)/3 = A049685(n-1),

a(5*n+1) = F(4*n-1) = A033891(n-1),

a(5*n+2) = L(4*n+2)/3 - F(4*n),

a(5*n+3) = L(4*n-2)/3 + F(4*n),

a(5*n+4) = F(4*n+1) = A033889(n). (End)