A276095 A nonlinear recurrence of order 4: a(1)=a(2)=a(3)=a(4)=1; a(n)=(a(n-1)+a(n-2)+a(n-3))^2/a(n-4).

1, 1, 1, 1, 9, 121, 17161, 298978681, 9933176210033041, 815437979830770470704295274609, 38747106750801481775941360512378545527545442200632960401

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. A072878, A072882.

Programs

Mathematica

RecurrenceTable[{a[n] == (a[n - 1] + a[n - 2] + a[n - 3])^2/a[n - 4], a[1] == a[2] == a[3] == a[4] == 1}, a, {n, 1, 12}] (* Michael De Vlieger, Aug 18 2016 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,(b+c+d)^2/a}; NestList[nxt,{1,1,1,1},10][[;;,1]] (* Harvey P. Dale, Aug 20 2024 *)

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 A276095(n)
  A(4, n)
end

Formula

a(n) = A072878(n)^2.

a(n) = 16*a(n-1)*a(n-2)*a(n-3) - 2a(n-1) - 2a(n-2) - 2a(n-3) - a(n-4).

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