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 1-10 of 18 results. Next

A006267 Continued cotangent for the golden ratio.

Original entry on oeis.org

1, 4, 76, 439204, 84722519070079276, 608130213374088941214747405817720942127490792974404
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A000032, A000204, A001622, A001999, A002666, A002667, A002668, A006266, A002813, A045529, A271223, A268924.

Programs

  • Maple
    a := proc(n) option remember; if n = 1 then 4 else a(n-1)^3 + 3*a(n-1) end if; end: seq(a(n), n = 1..5); # Peter Bala, Nov 15 2022
  • Mathematica
    c = N[GoldenRatio, 1000]; Table[Round[c^(3^n)], {n, 1, 8}] (* Artur Jasinski, Sep 22 2008 *)
a = {}; x = 4; Do[AppendTo[a, x]; x = x^3 + 3 x, {n, 1, 10}]; a (* Artur Jasinski, Sep 24 2008 *)
  • PARI
    a(n)=fibonacci(3^n+1) + fibonacci(3^n-1) \\ Andrew Howroyd, Dec 30 2024
  • PARI
    a(n)={my(t=1); for(i=1, n, t = t^3 + 3*t); t} \\ Andrew Howroyd, Dec 30 2024

Formula

(1+sqrt(5))/2 = cot(Sum_{n>=0} (-1)^n*acot(a(n))); let b(0) = (1+sqrt(5))/2, b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)). - Benoit Cloitre, Apr 10 2003
a(n) = A000204(3^n). - Benoit Cloitre, Sep 18 2005
a(n) = round(c^(3^n)) where c = GoldenRatio = 1.6180339887498948482... = (sqrt(5)+1)/2 (A001622). - Artur Jasinski, Sep 22 2008
a(n) = a(n-1)^3 + 3*a(n-1), a(0) = 1. - Artur Jasinski, Sep 24 2008
a(n+1) = Product_{k = 0..n} A002813(k). Thus a(n) divides a(n+1). - Peter Bala, Nov 22 2012
Sum_{n>=0} a(n)^2/A045529(n+1) = 1. - Amiram Eldar, Jan 12 2022
a(n) = Product_{k=0..n-1} (Lucas(2*3^k) + 1) (Usiskin, 1973). - Amiram Eldar, Jan 29 2022
From Peter Bala, Nov 15 2022: (Start)
a(n) = Lucas(3^n) for n >= 1.
a(n) == 1 (mod 3) for n >= 1.
a(n+1) == a(n) (mod 3^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).
The smallest positive residue of a(n) mod 3^n = A268924(n).
In the ring of 3-adic integers the limit_{n -> oo} a(n) exists and is equal to A271223. Cf. A006266. (End)

Extensions

The next term is too large to include.

A081782 Continued cotangent for the Gamma constant.

Original entry on oeis.org

0, 1, 3, 16, 389, 479403, 590817544217, 473341703003810973963339, 269963674630454468003021997747122421847127276823, 84255020180725066155718508782582560544360994462142096519461567461295107080386955008872752275165
Offset: 0

Views

Author

Benoit Cloitre, Apr 10 2003

Keywords

Links

Crossrefs

Cf. A001620, A002666, A002667.

Programs

  • Mathematica
    Floor[NestList[(#*Floor[#]+1)/(#-Floor[#]) &, EulerGamma, 9]] (* Stefano Spezia, Apr 24 2025 *)
  • PARI
    bn=vector(100);
b(n)=if(n<0,0,bn[n]);
bn[1]=Euler;
for(n=2,10,bn[n]=(b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))))
a(n)=floor(b(n+1))

Formula

Gamma = cot(Sum_{n>=0} (-1)^n*acot(a(n))).
Let b(0) = Gamma, b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)).

A081780 Continued cotangent for square root of 3.

Original entry on oeis.org

1, 3, 16, 405, 672320, 1352290629546, 4817801789405093582066881, 125242334906904794515426191021121381708761081802849, 48098205042808280233282792304806428848378744166690329294696649253234314988369597043479633480454760546
Offset: 0

Views

Author

Benoit Cloitre, Apr 10 2003

Keywords

Crossrefs

Cf. A002666, A002667.

Programs

  • Mathematica
    Floor[NestList[(#*Floor[#]+1)/(#-Floor[#]) &, Sqrt[3], 8]] (* Stefano Spezia, Apr 24 2025 *)
  • PARI
    \p500
bn=vector(100); bn[1]=sqrt(3); b(n)=if(n<0,0,bn[n]);
for(n=2,10,bn[n]=(b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))))
a(n)=floor(b(n+1))

Formula

sqrt(3) = cot(Sum_{n>=0} (-1)^n*acot(a(n))).
Let b(0) = sqrt(3), b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)).

A081781 Continued cotangent for square root of 5.

Original entry on oeis.org

2, 23, 2961, 9798690, 110528843365255, 16850689692769666225948933906, 357017084742873887104901208145335336566095925224005385073, 137046393407742688281781643318644289249627231296779665713406441509975367310318242383523303176564909084594247359836
Offset: 0

Views

Author

Benoit Cloitre, Apr 10 2003

Keywords

Crossrefs

Cf. A002666, A002667.

Programs

  • Mathematica
    Floor[NestList[(#*Floor[#]+1)/(#-Floor[#]) &, Sqrt[5], 7]] (* Stefano Spezia, Apr 24 2025 *)
  • PARI
    bn=vector(100);
b(n)=if(n<0,0,bn[n]);
bn[1]=sqrt(5);
for(n=2,10,bn[n]=(b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))));
a(n) = floor(b(n+1));

Formula

sqrt(5) = cot(Sum_{n>=0} (-1)^n*acot(a(n))).
Let b(0) = sqrt(5), b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)).

A081783 Continued cotangent for zeta(2) = Pi^2/6.

Original entry on oeis.org

1, 4, 172, 181307, 241328833528, 824652019956267685427678, 768422457901766762303892554138930904416139509281, 2110688056630901907060877896737932376507936264268382076456539236145849709148481095915090382331184
Offset: 0

Views

Author

Benoit Cloitre, Apr 10 2003

Keywords

Crossrefs

Cf. A001620, A002666, A002667.

Programs

  • PARI
    \p900
bn=vector(100);
bn[1]=Pi^2/6;
b(n)=if(n<0,0,bn[n]);
for(n=2,10,bn[n]=(b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))));
a(n)=floor(b(n+1));

Formula

Pi^2/6 = cot(Sum_{n>=0} (-1)^n*acot(a(n))).
Let b(0) = Pi^2/6, b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)).

A081784 Continued cotangent for zeta(3).

Original entry on oeis.org

1, 10, 122, 33429, 1447509608, 3251816299888840778, 10657606087425320549792856871886476385, 1233698091085791193532165615536619532897409600456434390187369062304735077655
Offset: 0

Views

Author

Benoit Cloitre, Apr 10 2003

Keywords

Crossrefs

Cf. A001620, A002666, A002667.

Programs

  • PARI
    \p900
bn=vector(100);
bn[1]=zeta(3);
b(n)=if(n<0,0,bn[n]);
for(n=2,10,bn[n]=(b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))));
a(n)=floor(b(n+1));

Formula

zeta(3) = cot(Sum_{n>=0} (-1)^n*acot(a(n))).
Let b(0) = zeta(3), b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)).

A081785 Continued cotangent for log(2).

Original entry on oeis.org

0, 1, 5, 55, 14187, 314681540, 254841004490568887, 116515298784769863036311841843812470, 16752745298264669840183664790312981823804061660537952218518280274360024
Offset: 0

Views

Author

Benoit Cloitre, Apr 10 2003

Keywords

Crossrefs

Cf. A001620, A002666, A002667.

Programs

  • PARI
    \p900
bn=vector(100);
bn[1]=log(2);
b(n)=if(n<0,0,bn[n]);
for(n=2,10,bn[n]=(b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))));
a(n)=floor(b(n+1));

Formula

log(2) = cot(Sum_{n>=0} (-1)^n*acot(a(n))).
Let b(0) = log(2), b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)).

A081786 Continued cotangent for log(3).

Original entry on oeis.org

1, 21, 1591, 5623795, 60586207387643, 11771746654268128293298294264, 223879928814222731378358322195036233470100990978181612407, 60370531495558553873551440308434816125184142701829318337696802178075590153432959525193229753905998692773283506988
Offset: 0

Views

Author

Benoit Cloitre, Apr 10 2003

Keywords

Crossrefs

Cf. A002666, A002667.

Programs

  • PARI
    \p900
bn=vector(100);
bn[1]=log(3);
b(n)=if(n<0,0,bn[n]);
for(n=2,10,bn[n]=(b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))));
a(n)=floor(b(n+1));

Formula

log(3) = cot(Sum_{n>=0} (-1)^n*acot(a(n))).
Let b(0) = log(3), b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)).

A081787 Continued cotangent for sqrt(e).

Original entry on oeis.org

1, 4, 208, 51198, 3265038057, 25300257957809599598, 1548008157389016603196793951803038609594, 15445738611564165990406534887324277271178568836676520360367688416251534382546319
Offset: 0

Views

Author

Benoit Cloitre, Apr 10 2003

Keywords

References

  • D. H. Lehmer, A cotangent analogue of continued fractions, Duke Math. J., 4 (1935), 323-340.

Crossrefs

Cf. A002666, A002667, A002668.

Programs

  • PARI
    \p900
bn=vector(100);
bn[1]=exp(1/2);
b(n)=if(n<0,0,bn[n]);
for(n=2,10,bn[n]=(b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))));
a(n)=floor(b(n+1));

Formula

sqrt(e) = cot(Sum_{n>=0} (-1)^n*acot(a(n))).
Let b(0) = sqrt(e), b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)).

A081788 Continued cotangent for sin(1).

Original entry on oeis.org

0, 1, 11, 209, 778615, 3961986619787, 108027609649678328362291208, 12797763868538691769539594849146740548395979750179143, 2398705889323117848234063941075093304477004809996203196876904292203062137833411276780250923333345577605421
Offset: 0

Views

Author

Benoit Cloitre, Apr 10 2003

Keywords

References

  • D. H. Lehmer, A cotangent analogue of continued fractions, Duke Math. J., 4 (1935), 323-340.

Crossrefs

Cf. A002666, A002667, A002668.

Programs

  • PARI
    \p900
bn=vector(100);
bn[1]=sin(1);
b(n)=if(n<0,0,bn[n]);
for(n=2,10,bn[n]=(b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))));
a(n)=floor(b(n+1));

Formula

sin(1) = cot(Sum_{n>=0} (-1)^n*acot(a(n))).
Let b(0) = sin(1), b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)).
Showing 1-10 of 18 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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