A072880 -id:A072880 - 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

A064098 a(n+1) = (a(n)^2 + a(n-1)^2)/a(n-2), with a(1) = a(2) = a(3) = 1.

Original entry on oeis.org

1, 1, 1, 2, 5, 29, 433, 37666, 48928105, 5528778008357, 811537892743746482789, 13460438563050022083842073547074914, 32770967840592833551621556305285371426044732591005957081

Offset: 1

Santi Spadaro, Sep 16 2001

nonn

This sequence was introduced by Dana Scott but possibly studied earlier by others. - James Propp, Jan 27 2005
Sequence gives the upper-left entries of the respective matrices
  [1 1] [1 0] [2 1] [5 2] [29 12] [433 179] [37666 15571]
  [1 2] [0 1] [1 1] [2 1] [12  5] [179  74] [15571  6437], ...
satisfying the recurrence M(n) = M(n-1) M(n-3)^(-1) M(n-1) (note that the Fibonacci numbers satisfy the additive version of this recurrence). - James Propp, Jan 27 2005
Define b(1) = b(2) = b(3) = 1; b(n) = (b(n-1) + b(n-2))^2/b(n-3); then a(n) = sqrt(b(n)). - Benoit Cloitre, Jul 28 2002
Any 3 successive terms of the sequence satisfy the Markov equation x^2 + y^2 + z^2 = 3 xyz. Therefore from the 3rd term on this is a subsequence of the Markov numbers, A002559. Also, we conjecture that the limit of log(log(a(n)))/n is log((sqrt(5) + 1)/2). - Martin Giese (martin.giese(AT)oeaw.ac.at), Oct 13 2005
A subsequence of the Markoff numbers A002559. - Andrew Hone, Jan 16 2006
The recursion exhibits the Laurent phenomenon. Let F(n) = Fibonacci(n), e(n) = F(n) - 1, a(1) = a1, a(2) = a2, a(3) = a3, a(n) = (a(n-1)^2 + a(n-3)^2) / a(n-3), b(n) = a(n) * a1^e(n-1) * a2^e(n-2) * a3^e(n-3). Then b(n) for n > 1 is an irreducible polynomial in {a1^2, a2^2, a3^2}, b(n) = (b(n-1)^2 + (b(n-2) * a1^F(n-4) * a2^F(n-5) * a3^F(n-6))^2) / b(n-3), and a(n) = a(n-1) * a(n-2) * (a1^2 + a2^2 + a3^2) / (a1 * a2 * a3) - a(n-3). - Michael Somos, Jan 12 2013
Starting with n = 5, a(n) is the largest number in row n - 5 of the Markov tree, A368546. These numbers are obtained by descending the tree in alternating right and left steps; their Farey indices (see A368546 for the definition) are ratios of successive Fibonacci numbers, 1/2, 2/3, 3/5, 5/8, ... See Aigner, Proposition 10.2. - Wouter Meeussen and William P. Orrick, Feb 11 2024

			G.f. = x + x^2 + x^3 + 2*x^4 + 5*x^5 + 29*x^6 + 433*x^7 + 37666*x^8 + ...

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

Harry J. Smith, Table of n, a(n) for n = 1..18
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].
S. Fomin and A. Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics, 28 (2002), 119-144.
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.
Andrew N. W. Hone, Growth of Mahler measure and algebraic entropy of dynamics with the Laurent property, arXiv:2109.08217 [math.NT], 2021.
KöMaL-Mathematical and Physical Journal for Secondary Schools, New advanced problems: proposed problem A265, April 2001.
L. J. Mordell, On the integer solutions of the equation x^2+y^2+z^2+2xyz=n, J. Lond. Math. Soc. 28 (1953), 500-510.
J. Propp, The combinatorics of frieze patterns and Markoff numbers, arXiv:math/0511633 [math.CO], 2005-2008.
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. A002559, A072878, A072879, A072880.

Markov tree: A327345, A368546.

[n le 3 select 1 else (Self(n-1)^2 + Self(n-2)^2)/Self(n-3): n in [1..16]]; // G. C. Greubel, Nov 07 2024

f:=proc(n) option remember; global K; local i;
if n <= K then 1
else add(f(n-i)^2,i=1..K-1)/f(n-K); fi; end;
K:=3;
[seq(f(n),n=1..10)]; # N. J. A. Sloane, Mar 17 2017

a[n_]:= (a[n-1]^2 +a[n-2]^2)/a[n-3]; a[1]=a[2]=a[3]=1; Array[a, 13] (* Or *)
a[n_]:= 3*a[n-1]*a[n-2] - a[n-3]; a[1]= a[2]= a[3]= 1; Array[a, 13] (* Robert G. Wilson v, Dec 26 2012 *)
nxt[{a_,b_,c_}]:={b,c,(c^2+b^2)/a}; NestList[nxt,{1,1,1},15][[;;,1]] (* Harvey P. Dale, Jul 07 2025 *)

{a(n) = if( n<1, n = 4-n); if( n<4, 1, 3 * a(n-1) * a(n-2) - a(n-3))}; /* Michael Somos, Jan 12 2013 */

{ a=a3=a2=a1=1; for (n = 1, 18, if (n>3, a=(a1^2 + a2^2)/a3; a3=a2; a2=a1; a1=a); write("b064098.txt", n, " ", a) ) } /* Harry J. Smith, Sep 06 2009 */

def A064098(n):
    def a(n): return 1 if n<4 else (a(n-1)^2 + a(n-2)^2)/a(n-3)
    return a(n)
[A064098(n) for n in range(16)] # G. C. Greubel, Nov 07 2024

Conjecture: lim_{n -> infinity} log(log(a(n)))/n exists = 0.48.... - Benoit Cloitre, Aug 07 2002. This is true - see below.
For this subsequence of the Markoff numbers, we have 2^(F(n-1) - 1) < a(n) < 3^(F(n-1) - 1) for n > 4, where F(n) are the Fibonacci numbers with F(0)=0, F(1)=1, F(n+1) = F(n) + F(n-1). Hence log(log(a(n)))/n tends to log((1 + sqrt(5))/2) as previously conjectured. - Andrew Hone, Jan 16 2006
a(n) = 3 * a(n-1) * a(n-2) - a(n-3). a(4-n) = a(n) for all n in Z. - Michael Somos, Jan 12 2013
a(n) ~ 1/3 * c^(((1 + sqrt(5))/2)^n), where c = 1.2807717799265504005186306582930649245... . - Vaclav Kotesovec, May 06 2015

Entry improved by comments from Michael Somos, Sep 25 2001

A022405 a(n) = a(n-1)*a(n-2) - a(n-3), with a(1) = 0, a(2) = 1, a(3) = 2.

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 10, 37, 366, 13532, 4952675, 67019597734, 331926286207224918, 22245566178948766568816183137, 7383888166355511098764350563784314022618210032

Offset: 1

Robert G. Wilson v, Jul 05 2000

Theorem 1.1 of Hare et al. (2010, 2011) involves a shifted version of this sequence and the Fibonacci sequence A000045. (The program by Alonso del Arte below does involve a shifted version of this sequence.) - Petros Hadjicostas, May 11 2019

G. C. Greubel, Table of n, a(n) for n = 1..21
Kevin G. Hare, Ian D. Morris, Nikita Sidorov, and Jacques Theys, An explicit counterexample to the Lagarias-Wang finiteness conjecture, arXiv:1006.2117 [math.OC], 2010-2011.
Kevin G. Hare, Ian D. Morris, Nikita Sidorov, and Jacques Theys, An explicit counterexample to the Lagarias-Wang finiteness conjecture, Advances in Mathematics 226 (2011), 4667-4701.

Cf. A001622, A061021, A061292, A072878, A072879, A072880, A178768.

I:=[0,1,2]; [n le 3 select I[n] else Self(n-1)*Self(n-2) - Self(n-3): n in [1..15]];  // G. C. Greubel, Mar 01 2018

a[1] = 0; a[2] = 1; a[3] = 2; a[n_] := a[n] = a[n - 1] a[n - 2] - a[n - 3]; Table[a[n], {n, 1, 15}] (* Alonso del Arte, Jan 31 2011 *)
nxt[{a_,b_,c_}]:={b,c,c*b-a}; NestList[nxt,{0,1,2},15][[;;,1]] (* Harvey P. Dale, Mar 23 2025 *)

It appears that lim_{n->infinity} log(a(n))/phi^n = 0.07743008049000107520747623421744398272089261907514..., where phi = (1 + sqrt(5))/2 is the golden ratio A001622. - Petros Hadjicostas and Jon E. Schoenfield, May 11 2019

Name clarified by Michel Marcus, May 10 2019

A061292 a(n) = a(n-1)a(n-2)a(n-3) - a(n-4) for n>3 with a(0) = a(1) = a(2) = a(3) = 2.

Original entry on oeis.org

2, 2, 2, 2, 6, 22, 262, 34582, 199330642, 1806032092550706, 12449434806576800059248920402, 4481765860945171681908664776799089162954814190172722

Offset: 0

Stephen G Penrice, Jun 04 2001

Any four consecutive terms are a solution to the Diophantine equation w^2 + x^2 + y^2 + z^2 = wxyz.

a(n) = 2 * A072878(n+1).

Vincenzo Librandi, Table of n, a(n) for n = 0..18
Michael Magee and Brady Haran, A simple equation that behaves weirdly, Numberphile, Youtube video, 2025.

Cf. A022405, A061021, A072878, A072879, A072880.

a061292 n = a061292_list !! n
a061292_list = 2 : 2 : 2 : 2 : zipWith (-)
   (zipWith3 (((*) .) . (*)) (drop 2 xs) (tail xs) xs) a061292_list
   where xs = tail a061292_list
-- Reinhard Zumkeller, Mar 25 2015

I:=[2,2,2,2]; [n le 4 select I[n] else Self(n-1)*Self(n-2)*Self(n-3)-Self(n-4): n in [1..12]]; // Vincenzo Librandi, Sep 17 2011

a[1] := 2; a[2] := 2; a[3] := 2; a[4] := 2; a[n_] := a[n - 1]*a[n - 2]*a[n - 3] - a[n - 4]; Table[a[n], {n, 1, 15}] (* Stefan Steinerberger, Mar 31 2006 *)
RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==2,a[n]==a[n-1]a[n-2]a[n-3]- a[n-4]},a[n],{n,12}] (* Harvey P. Dale, Sep 15 2011 *)

More terms from Larry Reeves (larryr(AT)acm.org) and Jason Earls, Jun 05 2001

A072877 a(1) = a(2) = a(3) = a(4) = 1; a(n) = (a(n-1)*a(n-3) + a(n-2)^4)/a(n-4).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 19, 119, 65339, 67258454, 959259994615659593, 171965197021698738644442682357, 12959040525296547835480490169418622922155526267774117749963303914461

Offset: 1

Benoit Cloitre, Jul 28 2002

A variation of a Somos-4 sequence with a(n-2)^4 in place of a(n-2)^2.

Seiichi Manyama, Table of n, a(n) for n = 1..17
Joshua Alman, Cesar Cuenca, and Jiaoyang Huang, Laurent phenomenon sequences, Journal of Algebraic Combinatorics 43(3) (2015), pp. 589-633.
S. Fomin and A. Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics 28 (2002), pp. 119-144.
David Gale, The strange and surprising saga of the Somos sequences, Math. Intelligencer 13(1) (1991), pp. 40-42.
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), pp. L171-L177.

Cf. A006720, A022405, A061292, A065918, A072878, A072879, A072880, A074394, A178768.

L[0]:=0; L[1]:=0; L[2]:=0; L[3]:=0; for n from 0 to 4000 do L[n+4]:=evalf(ln(1+exp(L[n+3]+L[n+1]-4*L[n+2]))+4*L[n+2]-L[n]): od: for n from 3990 to 4000 do print(evalf(ln(L[n+4])/(n+4))): od: #Note: L[n] is log(a[n]) # Andrew Hone, Nov 15 2005

nxt[{a_,b_,c_,d_}]:={b,c,d,(d*b+c^4)/a}; NestList[nxt,{1,1,1,1},15][[All,1]] (* Harvey P. Dale, Jun 01 2022 *)

Lim_{n->infinity} (log(log(a(n))))/n = log(2+sqrt(3))/2 = A065918/2 or about 0.658. - Andrew Hone, Nov 15 2005; corrected by Michel Marcus, May 12 2019
From Jon E. Schoenfield, May 12 2019: (Start)
It appears that, for n >= 1,
  a(n) = ceiling(e^(c0*x^n + d0/x^n)) if n is even,
         ceiling(e^(c1*x^n + d1/x^n)) if n is odd,
where
  x  = sqrt(2 + sqrt(3)) = (sqrt(2) + sqrt(6))/2
  c0 =   0.024915247166055931001426396817534982995670642690...
  c1 =   0.029604794868229453467890216788323427656809346011...
  d0 = -10.535089427608481105514469573411011428431309483956...
  d1 =  -2.856773870202800001336732759121362374871088274450...
(End)

Definition corrected by Matthew C. Russell, Apr 24 2012

A061021 a(n) = a(n-1)*a(n-2) - a(n-3) with a(0) = a(1) = a(2) = 3.

Original entry on oeis.org

3, 3, 3, 6, 15, 87, 1299, 112998, 146784315, 16586334025071, 2434613678231239448367, 40381315689150066251526220641224742, 98312903521778500654864668915856114278134197773017871243

Offset: 0

Stephen G Penrice, May 23 2001

Any three consecutive terms are a solution to the Diophantine equation x^2 + y^2 + z^2 = xyz.

Harry J. Smith, Table of n, a(n) for n = 0..17
Loren C. Larson, Solution to Problem Proposal 701, College Mathematics Journal 33 (2002), pp. 241-242.
Edward T. H. Wang, Problem Proposal 701, College Mathematics Journal 32 (2001), p. 211.

Cf. A022405, A061292, A072878, A072879, A072880, A074394, A178768.

a061021 n = a061021_list !! n
a061021_list = 3 : 3 : 3 : zipWith (-)
(tail $ zipWith (*) (tail a061021_list) a061021_list) a061021_list
-- Reinhard Zumkeller, Mar 25 2015

RecurrenceTable[{a[n] == a[n - 1] a[n - 2] - a[n - 3], a[0] == a[1] == a[2] == 3}, a, {n, 0, 12}] (* Michael De Vlieger, Aug 21 2016 *)

for (n=0, 17, if (n>2, a=a1*a2 - a3; a3=a2; a2=a1; a1=a, if (n==0, a=a3=3, if (n==1, a=a2=3, a=a1=3))); write("b061021.txt", n, " ", a)) \\ Harry J. Smith, Jul 16 2009

From Jon E. Schoenfield, May 12 2019: (Start)
It appears that, for n >= 1,
  a(n) = ceiling(e^(c0*phi^n - c1/(-phi)^n))
where
  phi = (1 + sqrt(5))/2,
   c0 = 0.4004033011137849744572073756789830081726425559860...
   c1 = 0.2798639753144007577581523025628820390768226527315...
(End)

More terms from Erich Friedman, Jun 03 2001

Name clarified by Petros Hadjicostas, May 11 2019

A074394 a(n) = a(n-1)*a(n-2) - a(n-3) with a(1) = 1, a(2) = 2, and a(3) = 3.

Original entry on oeis.org

1, 2, 3, 5, 13, 62, 801, 49649, 39768787, 1974480504962, 78522694637486171445, 155041529758800625329015665441303, 12174278697379026530632791354719900462885271361687873

Offset: 1

Henry Bottomley, Sep 24 2002

nonn

All consecutive quadruples are pairwise coprime. Multiples of 2 occur when n=2 mod 4, multiples of 3 when n=3 mod 4, multiples of 5 when n=4 mod 7, multiples of 7 when n=10 mod 14, multiples of 9 when n=7 or 11 mod 24, multiples of 10 when n=18 mod 28. Multiples of 4, 6 and 8 never occur.

			a(6) = a(5)*a(4) - a(3) = 13*5 - 3 = 62.

Reinhard Zumkeller, Table of n, a(n) for n = 1..19

Cf. A001622, A022405, A061021, A072878, A072879, A072880.

a074394 n = a074394_list !! (n-1)
a074394_list = 1 : 2 : 3 : zipWith (-)
   (tail $ zipWith (*) (tail a074394_list) a074394_list) a074394_list
-- Reinhard Zumkeller, Mar 25 2015

nxt[{a_,b_,c_}]:={b,c,b*c-a}; NestList[nxt,{1,2,3},15][[All,1]] (* Harvey P. Dale, Jan 21 2023 *)

Lim_{n->infinity} a(n+1)/a(n)^phi = 1, where phi is the golden ratio (1+sqrt(5))/2 = A001622. - Benoit Cloitre, Sep 26 2002
From Jon E. Schoenfield, May 13 2019: (Start)
It appears that, for n >= 2,
  a(n) = ceiling(e^(c*phi^n - d/(-phi)^n))
where
  phi = (1 + sqrt(5))/2
    c = 0.230193077518834725477008740044380256486365499661...
    d = 0.067704372842879037264190305626317036100889750046...
(End)

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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A064098 a(n+1) = (a(n)^2 + a(n-1)^2)/a(n-2), with a(1) = a(2) = a(3) = 1.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

SageMath

Formula

Extensions

A022405 a(n) = a(n-1)*a(n-2) - a(n-3), with a(1) = 0, a(2) = 1, a(3) = 2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Formula

Extensions

A061292 a(n) = a(n-1)*a(n-2)*a(n-3) - a(n-4) for n>3 with a(0) = a(1) = a(2) = a(3) = 2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

Extensions

A072877 a(1) = a(2) = a(3) = a(4) = 1; a(n) = (a(n-1)*a(n-3) + a(n-2)^4)/a(n-4).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A061021 a(n) = a(n-1)*a(n-2) - a(n-3) with a(0) = a(1) = a(2) = 3.

Views

Author

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.

A061292 a(n) = a(n-1)a(n-2)a(n-3) - a(n-4) for n>3 with a(0) = a(1) = a(2) = a(3) = 2.