A276097 A nonlinear recurrence of order 5: a(1)=a(2)=a(3)=a(4)=a(5)=1; a(n)=(a(n-1)+a(n-2)+a(n-3)+a(n-4))^2/a(n-5).

1, 1, 1, 1, 1, 16, 361, 143641, 20741472361, 430214650013601071641, 11567790319010747187536221088708755344001, 370675271093071368960746074163948008803845834307486807769098691609909105887376

Offset: 1

Author

Seiichi Manyama, Aug 18 2016

Keywords

• nonn

Comments

All terms are perfect squares.

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..15

Crossrefs

Cf. A072879, A072882, A276095.

Programs

Mathematica

RecurrenceTable[{a[1] == a[2] == a[3] == a[4] == a[5] == 1, a[n] == (a[n-1] + a[n-2] + a[n-3] + a[n-4])^2 / a[n-5]}, a, {n, 15}] (* Vincenzo Librandi, Aug 21 2016 *)

Ruby

def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n
    i = a[1..-1].inject(:+)
    j = i * i
    break if j % a[0] > 0
    a = *a[1..-1], j / a[0]
    ary << a[0]
  end
  ary
end
def A276097(n)
  A(5, n)
end

Formula

a(n) = A072879(n)^2.

a(n) = 25*a(n-1)*a(n-2)*a(n-3)*a(n-4) - 2a(n-1) - 2a(n-2) - 2a(n-3) - 2a(n-4) - a(n-5).

a(n)*a(n-1)*a(n-2)*a(n-3)*a(n-4) = ((a(n) + a(n-1) + a(n-2) + a(n-3) + a(n-4))/5)^2.