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 11 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.

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)).

A081789 Continued cotangent for cos(1).

Original entry on oeis.org

0, 1, 3, 31, 1896, 4466398, 51545652359935, 7892797673015743066290382811, 79208317703482281896053478218775642522933780224074809198, 195824124605155340760338534459689354174423866004140103635128465652592661899008938733448659610261796713081609084855
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

  • Mathematica
    Floor[NestList[(#*Floor[#]+1)/(#-Floor[#]) &, Cos[1], 9]] (* Stefano Spezia, Apr 23 2025 *)
  • PARI
    bn=vector(100);
bn[1]=cos(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

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

A081790 Continued cotangent for tan(1).

Original entry on oeis.org

1, 4, 32, 1158, 1815746, 15716561494212, 1184500978807872650350593387, 5321879016477546178356935033926215638755808624425727229, 28586857373644233013728565794450100157386617152409721820238727067747604580786570816033645416762395120483912199
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
    bn=vector(100);
b(n)=if(n<0,0,bn[n]);
bn[1]=tan(1);
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

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

A081791 Continued cotangent for sinh(1).

Original entry on oeis.org

1, 12, 361, 6098638, 40523300374005, 2249947521757014011818803347, 23934989997298328927868650983263126279480229971908985849, 2858972935444266747494645602151444724302966656792002957565134720098726922500919331865545624811189334200411357440
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]=sinh(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

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

A081792 Continued cotangent for cosh(1).

Original entry on oeis.org

1, 4, 28, 898, 865865, 6558406221253, 369641727028862496144018420, 168218383805281752399017936550348552720479497871513674, 46139813370820669084709611625366168409170012365100187639338625228748249752136723842763775088752136299316085
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]=cosh(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

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

A081793 Continued cotangent for tanh(1).

Original entry on oeis.org

0, 1, 7, 135, 35445, 44465908998, 5112887721516309845621, 75234509360529020708450352828794956245887456, 5786575206590910267083400178061771765781639734324927167565054640197289842752623499343753
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]=tanh(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

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

A081794 Continued cotangent for Pi/4.

Original entry on oeis.org

0, 1, 8, 211, 114681, 118304381067, 14093169772574392414247, 233069007722838136376547872705625127588988391, 148096265277934997326846757550268707006396575812305676278686643630022889932579135326452726
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]=Pi/4;
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/4 = cot(Sum_{n>=0} (-1)^n*acot(a(n))).
Let b(0) = Pi/4, b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)).

A081795 Continued cotangent for Pi/3.

Original entry on oeis.org

1, 43, 4975, 87377992, 18385473430682423, 5186411232443302687031694765612941, 47469894147223278266560159220413635233953187522490823346090207081760
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
    \p1200
bn=vector(100);
bn[1]=Pi/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

Pi/3 = cot(Sum_{n>=0} (-1)^n*acot(a(n))).
Let b(0) = Pi/3, 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 11 results. Next

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

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