A276123 -id:A276123 - 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

A283329 a(n) = (1 + Sum_{j=1..K-1} a(n-j) + a(n-1)*a(n-K+1))/a(n-K) with a(1),...,a(K)=1, where K=4.

Original entry on oeis.org

1, 1, 1, 1, 5, 13, 33, 217, 617, 1633, 10813, 30805, 81601, 540401, 1539601, 4078401, 27009205, 76949213, 203838433, 1349919817, 3845921017, 10187843233, 67468981613, 192219101605, 509188323201, 3372099160801, 9607109159201, 25449228316801, 168537489058405

Offset: 1

N. J. A. Sloane, Mar 17 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..1770
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 also.
Index entries for linear recurrences with constant coefficients, signature (0,0,51,0,0,-51,0,0,1).

Cf. A276123, A283330.

def A(k, n)
  a = Array.new(k, 1)
  ary = [1]
  while ary.size < n
    j = (1..k - 1).inject(1){|s, i| s + a[-i]} + a[1] * a[-1]
    break if j % a[0] > 0
    a = *a[1..-1], j / a[0]
    ary << a[0]
  end
  ary
end
def A283329(n)
  A(4, n)
end # Seiichi Manyama, Mar 18 2017

From Seiichi Manyama, Mar 18 2017: (Start)
a(3*n)   = 3*a(3*n-1) - a(3*n-2) - 1,
a(3*n+1) = 3*a(3*n)   - a(3*n-1) - 1,
a(3*n+2) = 7*a(3*n+1) - a(3*n)   - 1. (End)
From Colin Barker, Nov 03 2020: (Start)
G.f.: x*(1 + x + x^2 - 50*x^3 - 46*x^4 - 38*x^5 + 33*x^6 + 13*x^7 + 5*x^8) / ((1 - x)*(1 + x + x^2)*(1 - 50*x^3 + x^6)).
a(n) = 51*a(n-3) - 51*a(n-6) + a(n-9).
(End)

More terms from Seiichi Manyama, Mar 17 2017

A283330 a(n) = (1 + Sum_{j=1..K-1} a(n-j) + a(n-1)*a(n-K+1))/a(n-K) with a(1),...,a(K)=1, where K=5.

Original entry on oeis.org

1, 1, 1, 1, 1, 6, 16, 41, 106, 806, 2311, 6126, 16066, 122401, 351136, 931006, 2441881, 18604041, 53370241, 141506681, 371149801, 2827691726, 8111925376, 21508084401, 56412327826, 429790538206, 1232959286791, 3269087322166, 8574302679706, 65325334115481

Offset: 1

N. J. A. Sloane, Mar 17 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..1838
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 also.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,153,0,0,0,-153,0,0,0,1).

Cf. A276123, A283329.

a[n_] := a[n] = If[n <= 5, 1, With[{m = If[Mod[n, 4] == 2, 8, 3]}, m a[n-1] - a[n-2] - 1]];
Array[a, 30] (* Jean-François Alcover, Nov 03 2020 *)

def A(k, n)
  a = Array.new(k, 1)
  ary = [1]
  while ary.size < n
    j = (1..k - 1).inject(1){|s, i| s + a[-i]} + a[1] * a[-1]
    break if j % a[0] > 0
    a = *a[1..-1], j / a[0]
    ary << a[0]
  end
  ary
end
def A283330(n)
  A(5, n)
end # Seiichi Manyama, Mar 18 2017

From Seiichi Manyama, Mar 18 2017: (Start)
a(4*n-1) = 3*a(4*n-2) - a(4*n-3) - 1,
a(4*n)   = 3*a(4*n-1) - a(4*n-2) - 1,
a(4*n+1) = 3*a(4*n)   - a(4*n-1) - 1,
a(4*n+2) = 8*a(4*n+1) - a(4*n)   - 1. (End)
From Colin Barker, Nov 03 2020: (Start)
G.f.: x*(1 + x + x^2 + x^3 - 152*x^4 - 147*x^5 - 137*x^6 - 112*x^7 + 106*x^8 + 41*x^9 + 16*x^10 + 6*x^11) / ((1 - x)*(1 + x)*(1 + x^2)*(1 - 152*x^4 + x^8)).
a(n) = 153*a(n-4) - 153*a(n-8) + a(n-12) for n>12.
(End)

More terms from Seiichi Manyama, Mar 17 2017

A276284 a(0) = a(1) = a(2) = a(3) = a(4) = 1; for n > 4, a(n) = ( a(n-1)+a(n-3)+1 )*( a(n-2)+a(n-4)+1 ) / a(n-5).

Original entry on oeis.org

1, 1, 1, 1, 1, 9, 33, 385, 13825, 5474305, 8430415841, 1398605982547209, 30625582893143965429313, 3098236789946633955987434183345281, 17332850039068891068793031113694107707268123637761

Offset: 0

Seiichi Manyama, Aug 27 2016

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..21

Cf. A276123.

RecurrenceTable[{a[n] == (a[n - 1] + a[n - 3] + 1) (a[n - 2] + a[n - 4] + 1)/a[n - 5], a[0] == a[1] == a[2] == a[3] == a[4] == 1}, a, {n, 0, 14}] (* Michael De Vlieger, Aug 27 2016 *)
nxt[{a_,b_,c_,d_,e_}]:={b,c,d,e,(e+c+1) (d+b+1)/a}; NestList[nxt,{1,1,1,1,1},15][[All,1]] (* Harvey P. Dale, Dec 14 2021 *)

def A(m, n)
  a = Array.new(2 * m + 1, 1)
  ary = [1]
  while ary.size < n + 1
    i = (1..m).inject(1){|s, i| s + a[2 * i - 1]} * (1..m).inject(1){|s, i| s + a[2 * i]}
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276284(n)
  A(2, n)
end

a(n) = (8-4*(-1)^n)*a(n-1)*a(n-3) - a(n-2) - a(n-4) - 1 for n>3.

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.

Original entry on oeis.org

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

Seiichi Manyama, Aug 29 2016

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).

Cf. A276123, A276175.

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

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

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

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)

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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A283329 a(n) = (1 + Sum_{j=1..K-1} a(n-j) + a(n-1)*a(n-K+1))/a(n-K) with a(1),...,a(K)=1, where K=4.

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A283330 a(n) = (1 + Sum_{j=1..K-1} a(n-j) + a(n-1)*a(n-K+1))/a(n-K) with a(1),...,a(K)=1, where K=5.

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Programs

Mathematica

Ruby

Formula

Extensions

A276284 a(0) = a(1) = a(2) = a(3) = a(4) = 1; for n > 4, a(n) = ( a(n-1)+a(n-3)+1 )*( a(n-2)+a(n-4)+1 ) / a(n-5).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Ruby

Formula

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.

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

PARI

Ruby

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.