A005246 -id:A005246 - cp's OEIS frontend

A208213 a(n)=(a(n-1)^3*a(n-2)^2+1)/a(n-3) with a(0)=a(1)=a(2)=1.

1, 1, 1, 2, 9, 2917, 1005227383127, 960336157066554685167071011848947459782832

Offset: 0

Matthew C. Russell, Apr 23 2012

nonn

This is the case a=2, b=3, y(0)=y(1)=y(2)=1 of the recurrence shown in the Example 3.2 of "The Laurent phenomenon" (see Link lines, p. 10).

Seiichi Manyama, Table of n, a(n) for n = 0..9
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A005246, A208207, A208209, A208214.

a:=proc(n) if n<3 then return 1: fi: return (a(n-1)^3*a(n-2)^2+1)/a(n-3): end: seq(a(i),i=0..10);

RecurrenceTable[{a[0]==a[1]==a[2]==1,a[n]==(a[n-1]^3 a[n-2]^2+1)/a[n-3]},a,{n,10}] (* Harvey P. Dale, Jan 24 2014 *)

From Vaclav Kotesovec, May 20 2015: (Start)
a(n) ~ c1^(d1^n) * c2^(d2^n) * c3^(d3^n), where
d1 = -0.834243184313921717115626125802356204078143759301838339196857934562...
d2 = 0.3433795689528896338577674315423659679880371604828202900379886914176...
d3 = 3.4908636153610320832578586942599902360901065988190180491588692431448...
are the roots of the equation d^3 + 1 = 3*d^2 + 2*d and
c1 = 0.8780803541847027315058502579763355822688533316057717751329965683549...
c2 = 0.4420233041946828357635108827822581168188691631054586381824944218534...
c3 = 1.0154140443448836210836588567949793209798883476847171784955774310427...
(End)

A321119 a(n) = ((1 - sqrt(3))^n + (1 + sqrt(3))^n)/2^floor((n - 1)/2); n-th row common denominator of A321118.

Original entry on oeis.org

4, 2, 8, 10, 28, 38, 104, 142, 388, 530, 1448, 1978, 5404, 7382, 20168, 27550, 75268, 102818, 280904, 383722, 1048348, 1432070, 3912488, 5344558, 14601604, 19946162, 54493928, 74440090, 203374108, 277814198, 759002504, 1036816702, 2832635908, 3869452610

Offset: 0

Franck Maminirina Ramaharo, Nov 01 2018

			a(0) = ((1 - sqrt(3))^0 + (1 + sqrt(3))^0)/2^floor((0 - 1)/2) = 2*(1 + 1) = 4.

Harold J. Ahlberg, Edwin N. Nilson and Joseph L. Walsh, The Theory of Splines and Their Applications, Academic Press, 1967. See p. 47, Table 2.5.2.

Encyclopedia of Mathematics, Quadrature formula
John C. Holladay, A smoothest curve approximation, Math. Comp. Vol. 11 (1957), 233-243.
Peter Köhler, On the weights of Sard's quadrature formulas, CALCOLO Vol. 25 (1988), 169-186.
Leroy F. Meyers and Arthur Sard, Best approximate integration formulas, J. Math. Phys. Vol. 29 (1950), 118-123.
Arthur Sard, Best approximate integration formulas; best approximation formulas, American Journal of Mathematics Vol. 71 (1949), 80-91.
Isaac J. Schoenberg, Spline interpolation and best quadrature formulae, Bull. Amer. Math. Soc. Vol. 70 (1964), 143-148.
Frans Schurer, On natural cubic splines, with an application to numerical integration formulae, EUT report. WSK, Dept. of Mathematics and Computing Science Vol. 70-WSK-04 (1970), 1-32.
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-1).

Cf. A002176 (common denominators of Cotesian numbers).
Cf. A321118, A321120.
Cf. A001834, A002530, A002531, A003500, A005246, A048788, A083336, A125905.

LinearRecurrence[{0, 4, 0, -1}, {4, 2, 8, 10}, 50]

a(n) := ((1 - sqrt(3))^n + (1 + sqrt(3))^n)/2^floor((n - 1)/2)$
makelist(ratsimp(a(n)), n, 0, 50);

a(n) = (((sqrt(2) - sqrt(6))/2)^n + ((sqrt(6) + sqrt(2))/2)^n)*((2 - sqrt(2))*(-1)^n + 2 + sqrt(2))/2.
a(-n) = (-1)^n*a(n).
a(n) = 2*A000034(n+1)*A002531(n).
a(2*n) = 2*A001834(n).
a(2*n+1) = 2*A003500(n).
a(n) = 4*a(n-2) - a(n-4) with a(0) = 4, a(1) = 2, a(2) = 8, a(3) = 10.
a(2*n+3) = a(2*n+1) + a(2*n+2).
a(2*n+2) = a(2*n) + 2*a(2*n+1).
G.f.: 2*(1 - x)*(2 + 3*x - x^2)/(1 - 4*x^2 + x^4).
E.g.f.: (1 + exp(-sqrt(6)*x))*((2 - sqrt(2))*exp(sqrt(2 - sqrt(3))*x) + (2 + sqrt(2))*exp(sqrt(2 + sqrt(3))*x))/2.
Lim_{n->infinity} a(2*n+1)/a(2*n) = (1 + sqrt(3))/2.

A072713 a(1)=a(2)=a(3)=a(4)=a(5)=1; for n>5, a(n)a(n-5) = a(n-1)a(n-2)a(n-3)a(n-4)+1.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 3, 7, 43, 1807, 815861, 147917502976, 1339566593057489572791, 6793440021984612817314824762112917427331, 607759339422199886496126580428414916308278553796099069562650354036190535151

Offset: 1

Benoit Cloitre, Aug 07 2002

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..19

Cf. A005246, A051786.

nxt[{a_,b_,c_,d_,e_}]:={b,c,d,e,(b*c*d*e+1)/a}; Join[{1,1,1,1}, Transpose[ NestList[nxt,{1,1,1,1,1},15]][[5]]] (* Harvey P. Dale, Oct 03 2012 *)

A208206 a(n)=(a(n-1)^2*a(n-2)+1)/a(n-3) with a(0)=a(1)=a(2)=1.

Original entry on oeis.org

1, 1, 1, 2, 5, 51, 6503, 431347892, 23724602128927104843, 37334625705205335653803036700733450756576803

Offset: 0

Matthew C. Russell, Apr 23 2012

nonn

This is the case a=1, b=2, y(0)=y(1)=y(2)=1 of the recurrence shown in the Example 3.2 of "The Laurent phenomenon" (see Link lines, p. 10).

The next term (a(10)) has 98 digits. - Harvey P. Dale, Oct 04 2014

Seiichi Manyama, Table of n, a(n) for n = 0..12
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A005246, A208202, A208207, A208209.

a:=proc(n) if n<3 then return 1: fi: return (a(n-1)^2*a(n-2)+1)/a(n-3): end: seq(a(i),i=0..10);

RecurrenceTable[{a[0]==a[1]==a[2]==1,a[n]==(a[n-1]^2 a[n-2]+1)/a[n-3]},a,{n,10}] (* Harvey P. Dale, Oct 04 2014 *)

From Vaclav Kotesovec, May 20 2015: (Start)
a(n) ~ c1^(d1^n) * c2^(d2^n) * c3^(d3^n), where
d1 = -0.80193773580483825247220463901489010233183832426371430010712484639886484...
d2 = 0.554958132087371191422194871006410481067288862470910089376025968205157535...
d3 = 2.246979603717467061050009768008479621264549461792804210731098878193707304...
are the roots of the equation d^3 + 1 = 2*d^2 + d and
c1 = 0.874335057499939749225491691816700793966151250175012051621456437468590379...
c2 = 0.402356411273897640287204171338236092104516307383060911032953286637247174...
c3 = 1.071117422488325114038954501945557033632156032599675833309484054582086570...
(End)

A208208 a(n)=(a(n-1)^4*a(n-2)+1)/a(n-3) with a(0)=a(1)=a(2)=1.

Original entry on oeis.org

1, 1, 1, 2, 17, 167043, 6618080569762280805809

Offset: 0

Matthew C. Russell, Apr 23 2012

nonn

This is the case a=1, b=4, y(0)=y(1)=y(2)=1 of the recurrence shown in the Example 3.2 of "The Laurent phenomenon" (see Link lines, p. 10).

Seiichi Manyama, Table of n, a(n) for n = 0..8
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A005246, A208207.

a:=proc(n) if n<3 then return 1: fi: return (a(n-1)^4*a(n-2)+1)/a(n-3): end: seq(a(i),i=0..10);

RecurrenceTable[{a[n] == (a[n - 1]^4*a[n - 2] + 1)/a[n - 3], a[0] == a[1] == a[2] == 1}, a, {n, 0, 7}] (* Michael De Vlieger, Mar 19 2017 *)

From Vaclav Kotesovec, May 20 2015: (Start)
a(n) ~ c1^(d1^n) * c2^(d2^n) * c3^(d3^n), where
d1 = -0.588363990685104156421284586508527584304318862407786509166141051262...
d2 = 0.4064206546327112651910488344937800073049991477253475806754539682375...
d3 = 4.1819433360523928912302357520147475769993197146824389284906870830246...
are the roots of the equation d^3 + 1 = 4*d^2 + d and
c1 = 0.8094826741348488413005600397911253102639462301397489110738060562305...
c2 = 0.5758908197062035276668941188013698534573120455706764136847247903030...
c3 = 1.0094396347013780675988108222508397688561313671701492219003321772184...
(End)

A208214 a(n)=(a(n-1)^3*a(n-2)^3+1)/a(n-3) with a(0)=a(1)=a(2)=1.

Original entry on oeis.org

1, 1, 1, 2, 9, 5833, 72339160083737, 8347449602301100278574002746114271427525770715131218

Offset: 0

Matthew C. Russell, Apr 23 2012

nonn

This is the case a=3, b=3, y(0)=y(1)=y(2)=1 of the recurrence shown in the Example 3.2 of "The Laurent phenomenon" (see Link lines, p. 10).

Seiichi Manyama, Table of n, a(n) for n = 0..9
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A005246, A208210, A208213.

a:=proc(n) if n<3 then return 1: fi: return (a(n-1)^3*a(n-2)^3+1)/a(n-3): end: seq(a(i),i=0..10);

RecurrenceTable[{a[n] == (a[n - 1]^3*a[n - 2]^3 + 1)/a[n - 3], a[0] == a[1] == a[2] == 1}, a, {n, 0, 7}] (* Michael De Vlieger, Mar 19 2017 *)

From Vaclav Kotesovec, May 20 2015: (Start)
a(n) ~ c1^(d1^n) * c2^(d2^n) * c3^(d3^n), where
d1 = -1
d2 = 2-sqrt(3) = 0.2679491924311227064725536584941276330571947461896193719...
d3 = 2+sqrt(3) = 3.7320508075688772935274463415058723669428052538103806280...
are the roots of the equation d^3 + 1 = 3*d^2 + 3*d and
c1 = 0.9085343342123995498629194372995408229585378171837724081842452659181...
c2 = 0.3811823487030541690662698257664022175009714305688428757048879374472...
c3 = 1.0119167333492916399265234093841995850496968884402785055210058839859...
(End)

A076737 Let u(1)=u(2)=u(3)=2, u(n)=(1+u(n-1)u(n-2))/u(n-3); then a(n) is the numerator of u(n).

Original entry on oeis.org

2, 2, 2, 5, 3, 17, 11, 65, 43, 257, 171, 1025, 683, 4097, 2731, 16385, 10923, 65537, 43691, 262145, 174763, 1048577, 699051, 4194305, 2796203, 16777217, 11184811, 67108865, 44739243, 268435457, 178956971, 1073741825, 715827883, 4294967297

Offset: 1

Benoit Cloitre, Nov 24 2002

Robert Israel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,5,0,-4).

Cf. A005246, A076736 (denominator of u(n)).

2,2,2,seq(2/3+(1/6)*2^k+(1/12)*(-1)^k*2^k+(1/3)*(-1)^k,k=4..50); # Robert Israel, Aug 10 2015
H := (n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -8):
a := n -> `if`(n < 5, [2, 2, 2, 5][n], H(n-2, irem(n, 2), 1/2)):
seq(simplify(a(n)), n=1..34); # Peter Luschny, Sep 03 2019

nxt[{a_,b_,c_}]:={b,c,(1+b c)/a}; NestList[nxt,{2,2,2},40][[All,1]]// Numerator (* Harvey P. Dale, Oct 31 2021 *)

For n>4, a(n) = 2^A028242(n-4)*u(n); u(2n) = 2^(n-1)+1/2^n hence a(2n) = 4^(n-1)+1.
From Michael Somos (via Benoit Cloitre), Nov 29 2002: (Start)
a(1)=a(2)=a(3)=2, a(n+2) = (1+2^n)/(1+2*(n mod 2)).
For k>=2, a(2k+1) = A001045(2k-1). (End)
Empirical g.f.: x*(4*x^6+x^4-5*x^3-8*x^2+2*x+2) / ((x-1)*(x+1)*(2*x-1)*(2*x+1)). - Colin Barker, Oct 14 2014
This follows from the Somos formula for a(n+2). - Robert Israel, Aug 10 2015
a(1)=a(2)=a(3)=2 and, for n>3, a(n) = denominator(1/2+6/(4+2^n)). - Gerry Martens, Aug 10 2015
a(n) = H(n - 2, n mod 2, 1/2) for n >= 5 where H(n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -8). - Peter Luschny, Sep 03 2019

A208205 a(n)=(a(n-1)*a(n-2)^5+1)/a(n-3) with a(0)=a(1)=a(2)=1.

Original entry on oeis.org

1, 1, 1, 2, 3, 97, 11786, 33736797423001, 79097781524295318019203322936641

Offset: 0

Matthew C. Russell, Apr 23 2012

nonn

This is the case a=5, b=1, y(0)=y(1)=y(2)=1 of the recurrence shown in the Example 3.2 of "The Laurent phenomenon" (see Link lines, p. 10)

The next term (a(9)) has 96 digits. - Harvey P. Dale, Sep 13 2022

Seiichi Manyama, Table of n, a(n) for n = 0..11
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A005246, A208204.

a:=proc(n) if n<3 then return 1: fi: return (a(n-1)*a(n-2)^5+1)/a(n-3): end: seq(a(i),i=1..10);

RecurrenceTable[{a[n] == (a[n - 1] a[n - 2]^5 + 1)/a[n - 3], a[0] == a[1] == a[2] == 1}, a, {n, 0, 8}] (* Michael De Vlieger, Mar 19 2017 *)
nxt[{a_,b_,c_}]:={b,c,(c*b^5+1)/a}; NestList[nxt,{1,1,1},10][[All,1]] (* Harvey P. Dale, Sep 13 2022 *)

From Vaclav Kotesovec, May 20 2015: (Start)
a(n) ~ c1^(d1^n) * c2^(d2^n) * c3^(d3^n), where
d1 = -1.903211925911553287485216224057094233775314050044332659604216582082...
d2 = 0.1939365664746304482560845569332033002552873106788960042162607290276...
d3 = 2.7092753594369228392291316671238909335200267393654366553879558530545...
are the roots of the equation d^3 + 1 = d^2 + 5*d and
c1 = 0.9741074409555962981370572554321352591111177638556227590517984272608...
c2 = 0.0499759123576461468686480770163694779918691526759585723897652462761...
c3 = 1.0272217210627198315132544386598971884129462517962425299212701250318...
(End)

A208212 a(n) = (a(n-1)^2*a(n-2)^5+1)/a(n-3) with a(0)=a(1)=a(2)=1.

Original entry on oeis.org

1, 1, 1, 2, 5, 801, 1002501563, 66276977238296815913344374183794

Offset: 0

Matthew C. Russell, Apr 23 2012

nonn

This is the case a=5, b=2, y(0)=y(1)=y(2)=1 of the recurrence shown in the Example 3.2 of "The Laurent phenomenon" (see Link lines, p. 10).

Seiichi Manyama, Table of n, a(n) for n = 0..9
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001); Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A005246, A208211.

a:=proc(n) if n<3 then return 1: fi: return (a(n-1)^2*a(n-2)^5+1)/a(n-3): end: seq(a(i),i=0..10);

a[n_] := a[n] = If[n <= 2, 1, (a[n - 1]^2*a[n - 2]^5 + 1)/a[n - 3]];
Table[a[n], {n, 0, 7}] (* Jean-François Alcover, Nov 25 2017 *)

From Vaclav Kotesovec, May 20 2015: (Start)
a(n) ~ c1^(d1^n) * c2^(d2^n) * c3^(d3^n), where
d1 = -1.575773472651936072015953246349296378313356749177416595434978648425...
d2 = 0.1872837251102239188569922313039458439968721185362219238420888422761...
d3 = 3.3884897475417121531589610150453505343164846306411946715928898061494...
are the roots of the equation d^3 + 1 = 2*d^2 + 5*d and
c1 = 0.9607631794694254165284953988161129828633931861764073755339129251426...
c2 = 0.1625672201779811599302887070429221376610589038410298300467412998556...
c3 = 1.0141969317515019907302101637404918873873074910913934972790303073225...
(End)

A217787 a(n) = (a(n-1)*a(n-3) + 1) / a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 4, 9, 14, 19, 43, 67, 91, 206, 321, 436, 987, 1538, 2089, 4729, 7369, 10009, 22658, 35307, 47956, 108561, 169166, 229771, 520147, 810523, 1100899, 2492174, 3883449, 5274724, 11940723, 18606722, 25272721, 57211441, 89150161, 121088881

Offset: 0

Michael Somos, Mar 25 2013

This sequence is similar to A005246 whose recursion is a(n) = (a(n-1)*a(n-2) + 1) / a(n-3). - Michael Somos, Feb 10 2017

			G.f. = 1 + x + x^2 + x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 9*x^7 + 14*x^8 + 19*x^9 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..4411
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 (better) also. See Example 6.2.18.
Index entries for linear recurrences with constant coefficients, signature (0,0,5,0,0,-1).

Cf. A003501.

Cf. A005246, A048736, A006720, A072876, A072877, A111459.

[n le 3 select 1 else (Self(n)*Self(n-2)+1)/Self(n-3): n in [0..40]]; // Bruno Berselli, Mar 25 2013

a[ n_] := With[{m = If [n < 0, 3 - n, n]}, SeriesCoefficient[ (1 + x + x^2 - 4 x^3 - 3 x^4 - 2 x^5) / (1 - 5 x^3 + x^6), {x, 0, m}]]; (* Michael Somos, Jan 18 2015 *)
LinearRecurrence[{0,0,5,0,0,-1},{1,1,1,1,2,3},40] (* Harvey P. Dale, Nov 20 2016 *)

{a(n) = if( n<0, n = 3-n); polcoeff( (1 + x + x^2 - 4*x^3 - 3*x^4 - 2*x^5) / (1 - 5*x^3 + x^6) + x * O(x^n), n)};

G.f.: (1 + x + x^2 - 4*x^3 - 3*x^4 - 2*x^5) / (1 - 5*x^3 + x^6).
a(n) = a(3-n) for all n in Z.
a(n+3) + a(n-3) = 5*a(n) for all n in Z.
a(n+1) + a(n-1) = a(n) * (2 + [n mod 3 == 0]) for all n in Z.
a(n+3k)+a(n-3k) = A003501(k)*a(n) for n>=3k. - Bruno Berselli, Mar 25 2013

cp's OEIS Frontend

A208213 a(n)=(a(n-1)^3*a(n-2)^2+1)/a(n-3) with a(0)=a(1)=a(2)=1.

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A321119 a(n) = ((1 - sqrt(3))^n + (1 + sqrt(3))^n)/2^floor((n - 1)/2); n-th row common denominator of A321118.

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A072713 a(1)=a(2)=a(3)=a(4)=a(5)=1; for n>5, a(n)*a(n-5) = a(n-1)*a(n-2)*a(n-3)*a(n-4)+1.

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A208206 a(n)=(a(n-1)^2*a(n-2)+1)/a(n-3) with a(0)=a(1)=a(2)=1.

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A208208 a(n)=(a(n-1)^4*a(n-2)+1)/a(n-3) with a(0)=a(1)=a(2)=1.

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A208214 a(n)=(a(n-1)^3*a(n-2)^3+1)/a(n-3) with a(0)=a(1)=a(2)=1.

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A076737 Let u(1)=u(2)=u(3)=2, u(n)=(1+u(n-1)u(n-2))/u(n-3); then a(n) is the numerator of u(n).

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A208205 a(n)=(a(n-1)*a(n-2)^5+1)/a(n-3) with a(0)=a(1)=a(2)=1.

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A072713 a(1)=a(2)=a(3)=a(4)=a(5)=1; for n>5, a(n)a(n-5) = a(n-1)a(n-2)a(n-3)a(n-4)+1.