A289917 -id:A289917 - cp's OEIS frontend

A276175 a(n) = (a(n-1)+1)(a(n-2)+1)(a(n-3)+1)/a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.

1, 1, 1, 1, 8, 36, 666, 222111, 685187756, 2819713283228248, 644335913093223286486628176, 5604757351123068775966272886689217889936356651, 14861563788248216173988661093334637018340529129342104300621091389266132702213641

Offset: 0

Bruno Langlois, Aug 23 2016

nonn

Conjecture: a(n) is an integer for all n >= 0. It has been checked by computer for n <= 40. A proof was proposed by 'mercio' as an answer to the MSE question, which however lacks details and heavily relies on computation. [Updated by Max Alekseyev, May 07 2023]

Seiichi Manyama, Table of n, a(n) for n = 0..16
Math.StackExchange.com, Is A276175 integer-only? Aug 27 2016.

Cf. A076839, A276123, A362884.

RecurrenceTable[{a[n] == (a[n - 1] + 1) (a[n - 2] + 1) (a[n - 3] + 1)/a[n - 4],
a[0] == a[1] == a[2] == a[3] == 1}, a, {n, 0, 12}] (* Michael De Vlieger, Aug 25 2016 *)
a[ n_] := With[{m = Max[3 - n, n]}, If[ m < 4, 1, (a[m - 1] + 1) (a[m - 2] + 1) (a[m - 3] + 1)/a[m - 4]]]; (* Michael Somos, Jun 02 2019 *)

a(n) = if (n <=3, 1, (a(n-1)+1)*(a(n-2)+1)*(a(n-3)+1)/a(n-4)); \\ Michel Marcus, Aug 23 2016

def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[1..-1].inject(1){|s, i| s * (i + 1)}
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276175(n)
  A(4, n)
end # Seiichi Manyama, Aug 23 2016

Let b(n) = (b(n-1)*b(n-2)*b(n-3)+1)/b(n-4) with b(0) = 1/2, b(1) = 4, b(2) = b(3) = 1/2, then a(n) = b(n)*b(n+1)*b(n+2). - Seiichi Manyama, Sep 03 2016
The sequence 4*b(n) is given by A362884. Correspondingly, a(n) = A362884(n) * A362884(n+1) * A362884(n+2) / 64. - Max Alekseyev, May 07 2023
a(n) = a(3-n), 0 = a(n)*a(n+4)*(a(n+4)+1) - a(n+5)*a(n+1)*(a(n+1)+1) for all n in Z. - Michael Somos, Feb 23 2019
log(a(n)) ~ c * A289917^n, where c = 0.26774381278698... - Vaclav Kotesovec, Aug 27 2021

A289916 Coefficients of 1/(Sum_{k>=0} round((k+1)*r)(-x)^k), where r = 9/7.

Original entry on oeis.org

1, 3, 5, 8, 13, 22, 39, 69, 120, 206, 353, 607, 1046, 1803, 3106, 5348, 9208, 15856, 27306, 47025, 80982, 139457, 240155, 413566, 712196, 1226463, 2112073, 3637166, 6263503, 10786276, 18574872, 31987488, 55085136, 94861220, 163358969, 281317834, 484452887

Offset: 0

Clark Kimberling, Jul 18 2017

Conjecture: the sequence is strictly increasing.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-1,2,-1).

Cf. A078140 (includes guide to related sequences), A289917.

z = 2000; r = 9/7;
u = CoefficientList[Series[1/Sum[Round[(k + 1)*r] (-x)^k, {k, 0, z}], {x, 0, z}],
  x];  (* A289916  *)
v = N[u[[z]]/u[[z - 1]], 200]
RealDigits[v, 10][[1]] (* A289917 *)

Vec((1+x)^2*(1-x+x^2-x^3+x^4-x^5+x^6) / ((1-x+x^2)*(1-x-x^2-x^3+x^4)) + O(x^50)) \\ Colin Barker, Jul 20 2017

G.f.:  1/(Sum_{k>=0} round((k+1)*r)(-x)^k), where r = 9/7.
From Colin Barker, Jul 19 2017: (Start)
G.f.: (1+x)^2*(1-x+x^2-x^3+x^4-x^5+x^6) / ((1-x+x^2)*(1-x-x^2-x^3+x^4)).
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - a(n-4) + 2*a(n-5) - a(n-6) for n>5.
(End)

cp's OEIS Frontend

A276175 a(n) = (a(n-1)+1)(a(n-2)+1)(a(n-3)+1)/a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.

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A289916 Coefficients of 1/(Sum_{k>=0} round((k+1)*r)(-x)^k), where r = 9/7.

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cp's OEIS Frontend

A276175 a(n) = (a(n-1)+1)*(a(n-2)+1)*(a(n-3)+1)/a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Ruby

Formula

A289916 Coefficients of 1/(Sum_{k>=0} round((k+1)*r)(-x)^k), where r = 9/7.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A276175 a(n) = (a(n-1)+1)(a(n-2)+1)(a(n-3)+1)/a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.