A276308 a(n) = (a(n-1)+1)*(a(n-3)+1)/a(n-4) for n > 3, a(0) = a(1) = a(2) = a(3) = 1.

1, 1, 1, 1, 4, 10, 22, 115, 319, 736, 3886, 10816, 24991, 131989, 367405, 848947, 4483720, 12480934, 28839196, 152314471, 423984331, 979683706, 5174208274, 14402986300, 33280406797, 175770766825, 489277549849, 1130554147381, 5971031863756, 16621033708546

Offset: 0

Author

Seiichi Manyama, Aug 29 2016

Keywords

• nonn
• easy

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• S. Fomin and A. Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics, 28 (2002), 119-144.
• Matthew Christopher Russell, Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences, PhD Dissertation, Mathematics Department, Rutgers University, May 2016. See Eq. (6.137).
• Index entries for linear recurrences with constant coefficients, signature (0,0,35,0,0,-35,0,0,1).

Crossrefs

Cf. A276123, A276175.

Programs

PARI

Vec((1+x+x^2-34*x^3-31*x^4-25*x^5+22*x^6+10*x^7+4*x^8)/((1-x)*(1+x+x^2)*(1-34*x^3+x^6)) + O(x^35)) \\ Colin Barker, Aug 29 2016

PARI

a276308(maxn) = {a=vector(maxn); a[1]=a[2]=a[3]=a[4]=1; for(n=5, maxn, a[n]=(a[n-1]+1)*(a[n-3]+1)/a[n-4]); a} \\ Colin Barker, Aug 30 2016

Ruby

def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n + 1
    i = (a[1] + 1) * (a[-1] + 1)
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276308(n)
  A(4, n)
end

Formula

From Colin Barker, Aug 29 2016: (Start)

a(n) = 35*a(n-3)-35*a(n-6)+a(n-9) for n>8.

G.f.: (1+x+x^2-34*x^3-31*x^4-25*x^5+22*x^6+10*x^7+4*x^8) / ((1-x)*(1+x+x^2)*(1-34*x^3+x^6)).

(End)