A276531 a(n) = (a(n-1) * a(n-5) + a(n-2) * a(n-3) * a(n-4)) / a(n-6), with a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = 1.

1, 1, 1, 1, 1, 1, 2, 3, 5, 11, 41, 247, 1498, 39629, 3121233, 1344630757, 4527359876765, 673384475958949877, 12684198948982702826816701, 103442271685605704255863097581658042, 12389248756108266360505757651017660004796444483503, 657084395567781339286109602463271066924826185667810218784212689809097

Offset: 0

Author

Seiichi Manyama, Nov 16 2016

Keywords

• nonn

Comments

This sequence is the generalization of Dana Scott's sequence (A048736).

Conjecture: a(n) is an integer for all n. It has been checked by computer for 0 <= n <= 50.

The recursion has the Laurent property. If a(0), ..., a(5) are variables, then a(n) is a Laurent polynomial (a rational function with a monomial denominator). - Michael Somos, Nov 21 2016

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..28

Crossrefs

Cf. A048736, A006721.

Programs

GAP

a:=[1,1,1,1,1,1];; for n in [7..25] do a[n]:=(a[n-1]*a[n-5]+a[n-2]*a[n-3]*a[n-4])/a[n-6]; od; a; # Muniru A Asiru, Jul 30 2018

Magma

I:=[1,1,1,1,1,1]; [n le 6 select I[n] else (Self(n-1)*Self(n-5) + Self(n-2)*Self(n-3)*Self(n-4))/Self(n-6): n in [1..30]]; // G. C. Greubel, Jul 30 2018

Mathematica

RecurrenceTable[{a[n] == (a[n - 1] a[n - 5] + a[n - 2] a[n - 3] a[n - 4])/a[n - 6], a[0] == a[1] == a[2] == a[3] == a[4] == a[5] == 1}, a, {n, 0, 21}] (* Michael De Vlieger, Nov 21 2016 *)
nxt[{a_,b_,c_,d_,e_,f_}]:={b,c,d,e,f,(b*f+d*e*c)/a}; NestList[nxt,{1,1,1,1,1,1},30][[All,1]] (* Harvey P. Dale, Nov 21 2021 *)

Ruby

def A(k, n)
  a = Array.new(k, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[-1] * a[1] + a[2..-2].inject(:*)
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276531(n)
  A(6, n)
end

Formula

a(n) * a(n-6) = a(n-1) * a(n-5) + a(n-2) * a(n-3) * a(n-4).

a(5-n) = a(n) for all n in Z.