A064098 -id:A064098 - cp's OEIS frontend

A072878 a(n) = 4a(n-1)a(n-2)*a(n-3) - a(n-4) with a(1) = a(2) = a(3) = a(4) = 1.

1, 1, 1, 1, 3, 11, 131, 17291, 99665321, 903016046275353, 6224717403288400029624460201, 2240882930472585840954332388399544581477407095086361, 50384188378657848181032338163962292285660644698840136656562636145266593550842871302412156442811

Offset: 1

Benoit Cloitre, Jul 28 2002

A subsequence of the generalized Markoff numbers.

Seiichi Manyama, Table of n, a(n) for n = 1..16
Arthur Baragar, Integral solutions of the Markoff-Hurwitz equations, J. Number Theory 49 (1994), 27-44.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, arXiv:math/0601324 [math.NT], 2006.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, J. Phys. A: Math. Gen. 39 (2006), L171-L177.
Matthew Christopher Russell, Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrence, PhD Dissertation, Mathematics Department, Rutgers University, May 2016.

Cf. A022405, A064098, A075276, A072879, A072880.

RecurrenceTable[{a[1]==a[2]==a[3]==a[4]==1,a[n]==4a[n-1]a[n-2]a[n-3]-a[n-4]},a,{n,15}] (* Harvey P. Dale, Nov 29 2014 *)

a(1) = a(2) = a(3) = a(4) = 1; a(n) = (a(n-1)^2 + a(n-3)^2 + a(n-2)^2)/a(n-4) for n >= 5.

From the recurrence a(n) = 4*a(n-1)*a(n-2)*a(n-3) - a(n-4), any four successive terms satisfy the Markoff-Hurwitz equation a(n)^2 + a(n-1)^2 + a(n-2)^2 + a(n-3)^2 = 4*a(n)*a(n-1)*a(n-2)*a(n-3), cf. A075276. As n tends to infinity, the limit of log(log(a(n)))/n is log x = 0.6093778633..., where x=1.839286755... is the real root of the cubic x^3 - x^2 - x - 1 = 0. - Andrew Hone, Nov 14 2005

Entry revised Nov 19 2005, based on comments from Andrew Hone
a(13) from Harvey P. Dale, Nov 29 2014
Name clarified by Petros Hadjicostas, May 11 2019

A072879 a(n) = 5a(n-1)a(n-2)a(n-3)a(n-4) - a(n-5) with a(1) = a(2) = a(3) = a(4) = a(5) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 19, 379, 144019, 20741616379, 107553662508585672001, 608831069421618273050865038881215685876, 978035016076705458999330010986670207956236476587064788804921180339451725001

Offset: 1

Benoit Cloitre, Jul 28 2002

Solutions of the Hurwitz equation in five variables.

Seiichi Manyama, Table of n, a(n) for n = 1..16
Arthur Baragar, Integral solutions of the Markoff-Hurwitz equations, J. Number Theory 49 (1994), 27-44.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, arXiv:math/0601324 [math.NT], 2006.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, J. Phys. A: Math. Gen. 39 (2006), L171-L177.
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.

Cf. A006720, A064098, A072878, A072880.

nxt[{a_,b_,c_,d_,e_}]:={b,c,d,e,(5b c d e)-a}; NestList[nxt,{1,1,1,1,1},20][[All,1]] (* Harvey P. Dale, Nov 07 2016 *)

a(1) = a(2) = a(3) = a(4) = a(5) = 1; a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-4)^2)/a(n-5) for n >= 6.

From the recurrence a(n) = 5*a(n-1)*a(n-2)*a(n-3)*a(n-4) - a(n-5), any five successive terms satisfy the five-variable Hurwitz equation a(n)^2+a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-4)^2 = 5*a(n)*a(n-1)*a(n-2)*a(n-3)*a(n-4). As n tends to infinity, the limit of log(log(a(n)))/n is log x = 0.6562559790..., where x=1.927561975... is the largest real root of the quartic x^4-x^3-x^2-x-1=0. - Andrew Hone, Nov 16 2005

Entry revised Nov 19 2005, based on comments from Andrew Hone

Name clarified by Petros Hadjicostas, May 11 2019

A072880 A recurrence of order 6: a(1)=a(2)=a(3)=a(4)=a(5)=a(6)=1; a(n) = (a(n-1)^2 + a(n-2)^2 + a(n-3)^2 + a(n-4)^2 + a(n-5)^2)/a(n-6).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 5, 29, 869, 756029, 571580604869, 326704387862983487112029, 21347151409785350408171299054974277225256721769, 15713823217665540462976624783900822313284439536736221766688609460305249837839107387688348185

Offset: 1

Benoit Cloitre, Jul 28 2002

Any six successive terms satisfy the Markoff-Hurwitz equation a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6*a*b*c*d*e*f. - Bruno Langlois, Aug 09 2016

Seiichi Manyama, Table of n, a(n) for n = 1..17
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, arXiv:math/0601324 [math.NT], 2006.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, J. Phys. A: Math. Gen. 39 (2006), L171-L177.
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.

Cf. A006720, A064098, A072878, A072879.

RecurrenceTable[{a[n] == (a[n - 1]^2 + a[n - 2]^2 + a[n - 3]^2 + a[n -  4]^2 + a[n - 5]^2)/a[n - 6], a[1] == a[2] == a[3] == a[4] == a[5] == a[6] == 1}, a, {n, 1, 14}] (* Michael De Vlieger, Aug 11 2016 *)
nxt[{a_, b_, c_, d_, e_, f_}] := {b, c, d, e, f,(b^2+c^2+d^2+e^2+f^2)/a}; NestList[ nxt, Table[1,6],20][[All,1]] (* Harvey P. Dale, Mar 18 2018 *)

a(n) = 6*a(n-1)*a(n-2)*a(n-3)*a(n-4)*a(n-5) - a(n-6). - Bruno Langlois, Aug 09 2016

A368546 Alternative version of the Markov tree A327345.

Original entry on oeis.org

5, 13, 29, 34, 194, 433, 169, 89, 1325, 7561, 2897, 6466, 37666, 14701, 985, 233, 9077, 135137, 51641, 294685, 4400489, 1686049, 43261, 96557, 8399329, 48928105, 3276509, 1278818, 7453378, 499393, 5741, 610, 62210, 2423525, 925765, 13782649, 537169541

Offset: 0

William P. Orrick, Jan 04 2024

The Markov tree is a complete, infinite binary tree. Vertices are labeled by triples. The root vertex is (1, 5, 2). The left child of (a, b, c) is (a, 3*a*b - c, b); its right child is (b, 3*b*c - a, c). The sequence is a triangle read by rows consisting of the middle element of each triple, which is always the largest element of the triple. Row r contains 2^r elements.
The tree contains contains exactly one representative of each class of permutation equivalent nonsingular solutions to Markov's equation, a^2 + b^2 + c^2 = 3 * a * b * c. Nonsingular solutions are those in which a, b, and c are three distinct numbers. The two singular triples (1, 1, 1) and (1, 2, 1) are omitted in this sequence.
A consequence of Markov's equation is that the recurrence for the tree may be reformulated as follows: the left child of (a, b, c) is (a, (a^2 + b^2) / c, b); its right child is (b, (b^2 + c^2) / a, c).
An open problem is to prove the uniqueness conjecture, which asserts that the largest element of a triple determines the other two.
Frobenius proposed assigning a rational number index in (0,1) to each vertex of the tree, and hence to each term in this sequence. This is the Farey index, obtained by assigning the triple (0/1, 1/2, 1/1) to the root vertex and using the following rules to assign triples to the rest of the tree: the vertex labeled (u/v, w/x, y/z) with w = u + u and x = v + z has left child (u/v, (u+w)/(v+x), w/x) and right child (w/x, (w+y)/(x+z), y/z). The Farey index is the center element of the triple. Each rational number in (0, 1) appears as the Farey index of exactly one vertex of the tree. The index of a(n) is A007305(n+2) / A007306(n+2).
A sequence of leftward steps in the tree produces odd-indexed Fibonacci numbers, A001519, which have Farey indices of the form 1 / n. A sequence of rightward steps in the tree produces odd-indexed Pell numbers, A001653, which have Farey indices of the form (n - 1) / n. A sequence of leftward steps followed by a single rightward step produces A350922, corresponding to Farey indices of the form 2 / (2 * n + 1). Alternating steps right, left, right, left, right, ... produces A064098, which corresponds to Farey indices of the form F(n) / F(n + 1), where F(n) is the n-th Fibonacci number.

			The initial levels of the tree are as follows. (See p. 47 of Aigner's book.)
                               (1,5,2)
             (1,13,5)                              (5,29,2)
   (1,34,13)         (13,194,5)         (5,433,29)             (29,169,2)
(1,        (34,     (13,     (194,    (5,       (433,       (29,       (169,
 89,        1325,    7561,    2897,    6466,     37666,      14701,     985,
 ,34)        13)      194)     5)       433)      29)         169)       2)

Martin Aigner, Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings. Springer, 2013. x+257 pp. ISBN: 978-3-319-00887-5; 978-3-319-00888-2 MR3098784.

John Tyler Rascoe, Rows n = 0..10, flattened
Martin Aigner, Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings, [archive.org copy of the book].
Robert A. Gore, Patterns Within the Markov Tree, arXiv:2506.04299 [math.GM], 2025.

Other presentations of the Markov numbers, Markov triples, or the Markov tree: A002559, A253809, A291694, A305313, A305314, A327345.
Subsequences in the Markov tree: A001519, A001653, A350922, A064098.
Farey tree: A007305, A007306.

def Mtree(x): return(x[0],(3*x[0]*x[1])-x[2],x[1]), (x[1],(3*x[1]*x[2])-x[0],x[2])
def A368546_rowlist(maxrow):
    A,B = [[(1,5,2)]],[]
    for n in range(maxrow+1):
        A.append([])
        for j in A[n]:
            B.append(max(j))
            for k in Mtree(j):
                A[n+1].append(k)
    return(B) # John Tyler Rascoe, Feb 09 2024

def stripUpToFirst1(w):
    x = w
    while x % 2 == 0:
        x = x // 2
    return(x // 2)
def stripUpToFirst0(w):
    x = w
    while x % 2 == 1:
        x = x // 2
    if x == 0:
        return(None)
    else:
        return(x // 2)
@CachedFunction
def markovNumber(w):
    if w == None:
        return(2)
    elif w == 0:
        return(1)
    elif w == 1:
        return(5)
    elif w % 2 == 0:
        return(3*markovNumber(stripUpToFirst1(w))*markovNumber(w//2) - markovNumber(stripUpToFirst0(w//2)))
    else:
        return(3*markovNumber(stripUpToFirst0(w))*markovNumber(w//2) - markovNumber(stripUpToFirst1(w//2)))
[markovNumber(w) for w in range(1,38)]

A165903 a(n) = (a(n-1)^2 + a(n-2)^2 + a(n-1)*a(n-2))/a(n-3) with three initial ones.

Original entry on oeis.org

1, 1, 1, 3, 13, 217, 16693, 21717363, 2175145909081, 283430597537694797281, 3699017428454717709381715649628841, 6290488320295607125006566146327310005599469877825552723

Offset: 0

Jaume Oliver Lafont, Sep 29 2009

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..17

Cf. A064098, A072882, A165896.

a:=[1,1,1];; for n in [4..12] do a[n]:= (a[n-1]^2 + a[n-2]^2 + a[n-1]*a[n-2])/a[n-3]; od; a; # G. C. Greubel, Dec 19 2019

I:=[1,1,1]; [n le 3 select I[n] else (Self(n-1)^2 + Self(n-2)^2 + Self(n-1)*Self(n-2))/Self(n-3): n in [1..12]]; // G. C. Greubel, Dec 19 2019

a:= proc(n, k) option remember;
      if n<3 then 1
    else (a(n-1)^2 + a(n-2)^2 + a(n-1)*a(n-2))/a(n-3)
      fi; end:
seq( a(n), n=0..12); # G. C. Greubel, Dec 19 2019

RecurrenceTable[{a[0]==1,a[1]==1,a[2]==1, a[n]==(a[n-1]^2+a[n-2]^2+a[n-1]*a[n-2])/a[n-3]},a,{n,0,10}] (* Vaclav Kotesovec, May 06 2015 *)
nxt[{a_,b_,c_}]:={b,c,(c^2+b^2+b*c)/a}; NestList[nxt,{1,1,1},10][[All,1]] (* Harvey P. Dale, Oct 24 2022 *)

a(n)=if(n<3,1,(a(n-1)^2 +a(n-2)^2 +a(n-1)*a(n-2))/a(n-3))

@CachedFunction
def a(n):
    if (n<3): return 1
    else: return (a(n-1)^2+a(n-2)^2+a(n-1)*a(n-2))/a(n-3)
[a(n) for n in (0..12)] # G. C. Greubel, Dec 19 2019

a(n) ~ 1/6 * c^(((1+sqrt(5))/2)^n), where c = 1.902254978346365075882696720546123493664... . - Vaclav Kotesovec, May 06 2015

a(n) = 6*a(n-1)*a(n-2)-a(n-1)-a(n-2)-a(n-3). - Bruno Langlois, Aug 21 2016

"frac" keyword removed by Jaume Oliver Lafont, Oct 13 2009

A072882 A nonlinear recurrence of order 3: a(1)=a(2)=a(3)=1; a(n)=(a(n-1)+a(n-2))^2/a(n-3).

Original entry on oeis.org

1, 1, 1, 4, 25, 841, 187489, 1418727556, 2393959458891025, 30567386265691995561839449, 658593751358960570203157512237008273218521, 181183406309644143341701434158730639946454023369335051404405528107396

Offset: 1

Benoit Cloitre, Jul 28 2002

All terms are perfect squares.

Seiichi Manyama, Table of n, a(n) for n = 1..17

Cf. A064098, A276095, A276097.

RecurrenceTable[{a[1]==1,a[2]==1,a[3]==1, a[n]==(a[n-1]+a[n-2])^2/a[n-3]},a,{n,1,10}] (* Vaclav Kotesovec, May 06 2015 *)

a(n) ~ 1/9 * c^(((1+sqrt(5))/2)^n), where c = 1.6403763522562240514693138664331346215549... . - Vaclav Kotesovec, May 06 2015
a(n) = A064098(n)^2. - Seiichi Manyama, Aug 18 2016
From Seiichi Manyama, Aug 26 2016: (Start)
a(n) = 9*a(n-1)*a(n-2) - 2*a(n-1) - 2*a(n-2) - a(n-3).
a(n)*a(n-1)*a(n-2) = ((a(n) + a(n-1) + a(n-2))/3)^2. (End)

A276122 a(0) = a(1) = a(2) = 1; for n > 2, a(n) = (a(n-1)^2+a(n-2)^2+a(n-1)+a(n-2))/a(n-3).

Original entry on oeis.org

1, 1, 1, 4, 22, 526, 69427, 219111589, 91273561736491, 119994570874632853695766, 65713991236617279734602790963627271046, 47311933073383646516067037755547920981262829886906923065810924

Offset: 0

Bruno Langlois, Aug 21 2016

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..16

Cf. A064098, A072881, A101879, A165903.

RecurrenceTable[{a[n] == (a[n - 1]^2 + a[n - 2]^2 + a[n - 1] + a[n - 2])/a[n - 3], a[0] == a[1] == a[2] == 1}, a, {n, 0, 11}] (* Michael De Vlieger, Aug 21 2016 *)

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

a(n) ~ 1/6 * c^(phi^n), where c = 2.059783590102273... and phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Mar 20 2017

a(10) corrected by Seiichi Manyama, Aug 21 2016

A276258 a(n) = 4a(n-1)a(n-2) - a(n-3), with a(1) = a(2) = a(3) = 1.

Original entry on oeis.org

1, 1, 1, 3, 11, 131, 5761, 3018753, 69564144001, 839987873581797251, 233732149587751710483796746251, 785328685279672432967483833110876164468741280003, 734226246973363127354668827312570246092792043625372932024478449584047744277761

Offset: 1

Seiichi Manyama, Aug 25 2016

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..18

Cf. A001519, A064098, A276256, A276259.

RecurrenceTable[{a[n] == 4*a[n - 1]*a[n - 2] - a[n - 3], a[1] == 1,
  a[2] == 1, a[3] == 1}, a, {n, 1, 10}] (* G. C. Greubel, Aug 25 2016 *)

def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n
    a = *a[1..-1], *a[1..-1].inject(:*) * (m + 1) - a[0]
    ary << a[0]
  end
  ary
end
def A276258(n)
  A(3, n)
end

a(1)=a(2)=a(3)=1; a(n)=(a(n-1)^2+a(n-2)^2+1)/a(n-3).
a(n) ~ 1/4 * c^(((1+sqrt(5))/2)^n), where c = 1.41452525081158447693692520473959... . - Vaclav Kotesovec, Aug 26 2016
a(n)*a(n+1)*a(n+2) = (a(n)^2+a(n+1)^2+a(n+2)^2+1)/4. - Seiichi Manyama, Sep 04 2016

A258161 a(n) = a(n-1)^3/a(n-2) with a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 8, 256, 2097152, 36028797018963968, 22300745198530623141535718272648361505980416

Offset: 0

Vaclav Kotesovec, May 22 2015

The next term has 114 digits.

Cf. A000045, A000301, A001906, A064098, A208207, A230900, A258169.

Clear[a]; RecurrenceTable[{a[n]==a[n-1]^3/a[n-2], a[0]==1, a[1]==2},a,{n,0,8}]
Clear[a]; a[0]=2; a[n_]:=a[n]=Product[a[j]^(n-j),{j,0,n-1}]; Flatten[{1, Table[a[n], {n,1,8}]}]
Table[2^(Fibonacci[2*n]), {n, 0, 8}]

a(n) = 2^(Fibonacci(2*n)).

cp's OEIS Frontend

A072878 a(n) = 4*a(n-1)*a(n-2)*a(n-3) - a(n-4) with a(1) = a(2) = a(3) = a(4) = 1.

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A072879 a(n) = 5*a(n-1)*a(n-2)*a(n-3)*a(n-4) - a(n-5) with a(1) = a(2) = a(3) = a(4) = a(5) = 1.

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A072880 A recurrence of order 6: a(1)=a(2)=a(3)=a(4)=a(5)=a(6)=1; a(n) = (a(n-1)^2 + a(n-2)^2 + a(n-3)^2 + a(n-4)^2 + a(n-5)^2)/a(n-6).

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A368546 Alternative version of the Markov tree A327345.

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A165903 a(n) = (a(n-1)^2 + a(n-2)^2 + a(n-1)*a(n-2))/a(n-3) with three initial ones.

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A072882 A nonlinear recurrence of order 3: a(1)=a(2)=a(3)=1; a(n)=(a(n-1)+a(n-2))^2/a(n-3).

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A276122 a(0) = a(1) = a(2) = 1; for n > 2, a(n) = (a(n-1)^2+a(n-2)^2+a(n-1)+a(n-2))/a(n-3).

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A072878 a(n) = 4a(n-1)a(n-2)*a(n-3) - a(n-4) with a(1) = a(2) = a(3) = a(4) = 1.

A072879 a(n) = 5a(n-1)a(n-2)a(n-3)a(n-4) - a(n-5) with a(1) = a(2) = a(3) = a(4) = a(5) = 1.

A276258 a(n) = 4a(n-1)a(n-2) - a(n-3), with a(1) = a(2) = a(3) = 1.