cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A208210 a(n)=(a(n-1)^2*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, 5, 201, 2525063, 10355298070412763074, 8589063344901709900442551790362661608528200120823830773
Offset: 0

Views

Author

Matthew C. Russell, Apr 23 2012

Keywords

Comments

This is the case a=3, 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(9) -- has 161 digits. - Harvey P. Dale, Apr 14 2022

Links

Crossrefs

Cf. A005246, A208203, A208209, A208211, A208214.

Programs

  • Maple
    a:=proc(n) if n<3 then return 1: fi: return (a(n-1)^2*a(n-2)^3+1)/a(n-3): end: seq(a(i),i=0..10);
  • Mathematica
    a[0] = a[1] = a[2] = 1; a[n_] := a[n] = (a[n-1]^2*a[n-2]^3 + 1)/a[n-3];
Array[a, 10, 0] (* Jean-François Alcover, Dec 14 2017 *)
nxt[{a_,b_,c_}]:={b,c,(c^2 b^3+1)/a}; NestList[nxt,{1,1,1},10][[All,1]] (* Harvey P. Dale, Apr 14 2022 *)

Formula

From Vaclav Kotesovec, May 20 2015: (Start)
a(n) ~ c1^(d1^n) * c2^(d2^n) * c3^(d3^n), where
d1 = -1.198691243515997113071999692569776193916276872472594369204332359716...
d2 = 0.2864620650316004980582127604312427653427138786836169481458128553091...
d3 = 2.9122291784843966150137869321385334285735629937889774210585195044073...
are the roots of the equation d^3 + 1 = 2*d^2 + 3*d and
c1 = 0.9326266928252752296152676800592959458631493222642463226349218269187...
c2 = 0.2535475214701961189033928082745089316567819534655391761010907360554...
c3 = 1.0248087086665041891835364490857429725941144848712661648932932629036...
(End)

A208202 a(n) = (a(n-1)*a(n-2)^2+1)/a(n-3) with a(0)=a(1)=a(2)=1.

Original entry on oeis.org

1, 1, 1, 2, 3, 13, 59, 3324, 890065, 166683166499, 39725939269090918399, 1240040687243304530118746458657221560, 11740660815927242416329935330676365456512664243108711550072429939
Offset: 0

Views

Author

Matthew C. Russell, Apr 23 2012

Keywords

Comments

This is the case a=2, 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).

Links

Crossrefs

Cf. A005246, A208203, A208209.

Programs

  • Magma
    [n le 3 select 1 else (Self(n-1)*Self(n-2)^2+1)/Self(n-3): n in [1..13]]; // Bruno Berselli, Apr 24 2012
  • Mathematica
    RecurrenceTable[{a[0] == a[1] == a[2] == 1, a[n] == (a[n - 1] a[n - 2]^2 + 1)/a[n - 3]}, a, {n, 12}] (* Bruno Berselli, Apr 25 2012 *)
(* The numerical values of the constants d1, d2, d3 *) Print[N[{Root[1-2*#1-#1^2+#1^3&,1], Root[1-2*#1-#1^2+#1^3&,2], Root[1-2*#1-#1^2+#1^3&,3]}, 80]]; (* and the constants c1, c2, c3 *) A208202 = RecurrenceTable[{a[0]==a[1]==a[2]==N[1,100], a[n] == (a[n-1]*a[n-2]^2 + 1)/a[n-3]}, a, {n,1,30}]; Table[Flatten[N[{Exp[cc1], Exp[cc2], Exp[cc3]}/.Solve[Table[Log[A208202[[n]]] == cc1*Root[1 - 2*#1 - #1^2 + #1^3&, 1]^n + cc2*Root[1 - 2*#1 - #1^2 + #1^3&, 2]^n + cc3*Root[1 - 2*#1 - #1^2 + #1^3&, 3]^n, {n, k, k+2}]],80]], {k, Length[A208202]-3, Length[A208202]-2}] (* Vaclav Kotesovec, May 20 2015 *)

Formula

From Vaclav Kotesovec, May 20 2015: (Start)
a(n) ~ c1^(d1^n) * c2^(d2^n) * c3^(d3^n), where
d1 = -1.24697960371746706105000976800847962126454946179280421073109887819...
d2 = 0.445041867912628808577805128993589518932711137529089910623974031794...
d3 = 1.801937735804838252472204639014890102331838324263714300107124846398...
are the roots of the equation d^3 + 1 = d^2 + 2*d and
c1 = 0.937508205283971584227188160392119895660526011507051773879367647962...
c2 = 0.127128212809518009874462927372545164747593272064601714573478901156...
c3 = 1.135040592200579625529345655593495454581148721169010026906480955795...
(End)

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

Original entry on oeis.org

1, 1, 1, 2, 9, 2917, 1005227383127, 960336157066554685167071011848947459782832
Offset: 0

Views

Author

Matthew C. Russell, Apr 23 2012

Keywords

Comments

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).

Links

Crossrefs

Cf. A005246, A208207, A208209, A208214.

Programs

  • Maple
    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);
  • Mathematica
    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 *)

Formula

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)

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

Views

Author

Matthew C. Russell, Apr 23 2012

Keywords

Comments

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

Links

Crossrefs

Cf. A005246, A208202, A208207, A208209.

Programs

  • Maple
    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);
  • Mathematica
    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 *)

Formula

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)

A276266 a(0) = a(1) = a(2) = 1; for n>2, a(n) = ( a(n-1)*a(n-2) + 1 )^2 / a(n-3).

Original entry on oeis.org

1, 1, 1, 4, 25, 10201, 16259565169, 1100432328310492581042546436, 31383529740086705883339675381564403354342372463018335778292540655564225
Offset: 0

Views

Author

Seiichi Manyama, Aug 26 2016

Keywords

Links

Crossrefs

Cf. A081068, A208209.

Programs

  • Mathematica
    RecurrenceTable[{a[n] == (a[n - 1] a[n - 2] + 1)^2/a[n - 3], a[0] == a[1] == a[2] == 1}, a, {n, 0, 8}] (* Michael De Vlieger, Aug 26 2016 *)
  • Ruby
    def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[1..-1].inject(:*) + 1
    i *= i
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276266(n)
  A(3, n)
end

Formula

a(n) = A208209(n)^2.

A276267 a(n) = ( a(n-1)^2*a(n-2)^2*a(n-3)^2 + 1 ) / a(n-4), with a(0)=a(1)=a(2)=a(3)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 5, 101, 1020101, 132690278976255013, 37379828474243017116309068570169440106423243719554
Offset: 0

Views

Author

Seiichi Manyama, Aug 26 2016

Keywords

Links

Crossrefs

Cf. A001519, A208209.

Programs

  • Mathematica
    RecurrenceTable[{a[n] == (a[n - 1]^2 a[n - 2]^2 a[n - 3]^2 + 1)/a[n - 4], a[0] == a[1] == a[2] == a[3] == 1}, a, {n, 0, 10}] (* Michael De Vlieger, Aug 26 2016 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,(b^2 c^2 d^2+1)/a}; NestList[nxt,{1,1,1,1},10][[All,1]] (* Harvey P. Dale, Nov 18 2021 *)
  • Ruby
    def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n + 1
    i = (a[1..-1].inject(:*)) ** 2 + 1
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276267(n)
  A(4, n)
end
Showing 1-6 of 6 results.

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