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.

Previous Showing 11-18 of 18 results.

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)).

A081796 Continued cotangent for sin(Pi/3) = sqrt(3)/2.

Original entry on oeis.org

0, 1, 13, 196, 257087, 249639161983, 553029809670900697241813, 575598315149214535162520163688459972096324096213, 680813056961507163626080261194823226597566577785481001106845521689287461487322891517719568410606
Offset: 0

Views

Author

Benoit Cloitre, Apr 10 2003

Keywords

References

  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 433-434.
  • D. H. Lehmer, A cotangent analogue of continued fractions, Duke Math. J., 4 (1935), 323-340.

Links

Crossrefs

Cf. A002065, A002666, A002667, A002668.

Programs

  • PARI
    \p900
bn=vector(100);
bn[1]=sqrt(3)/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(3)/2 = cot(Sum_{n>=0} (-1)^n*acot(a(n))).
Let b(0) = sqrt(3)/2, b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)).
Previous Showing 11-18 of 18 results.

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

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