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.

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A033090 Indices of incrementally largest terms in the continued fraction for Pi.

Original entry on oeis.org

1, 2, 3, 5, 308, 432, 28422, 156382, 267314, 453294, 11504931, 849955263, 2349980289, 3588031780, 8600404591, 15621034283
Offset: 1

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Author

Keywords

Comments

This sequence assumes nonstandard indexing of continued fraction terms as [a_1; a_2, a_3, ...]. If you use the actual offset from A001203, corresponding to [a_0; a_1, a_2, ...], you get instead 0, 1, 2, 4, 307, 431, 28421, ... Compare with A033092 versus A224849. - Jeppe Stig Nielsen, Dec 14 2019

Crossrefs

Programs

  • Mathematica
    With[{s = ContinuedFraction[Pi, 2*10^7]}, Map[FirstPosition[s, #][[1]] &, Union@ FoldList[Max, s]]] (* Michael De Vlieger, Jan 31 2020 *)

Extensions

a(12) from Eric W. Weisstein, Dec 08 2010
a(13) from Eric W. Weisstein, Sep 16 2011
a(14) from Eric W. Weisstein, Sep 17 2011
a(15) from Eric W. Weisstein, Jul 18 2013
a(16) from Syed Fahad, Apr 27 2021

A154883 Distinct entries in continued fraction for Pi in the order of their appearance.

Original entry on oeis.org

3, 7, 15, 1, 292, 2, 14, 84, 13, 4, 6, 99, 5, 8, 12, 16, 161, 45, 22, 24, 10, 26, 42, 9, 57, 18, 19, 30, 28, 20, 120, 23, 21, 127, 29, 11, 48, 436, 58, 34, 44, 20776, 94, 55, 32, 50, 43, 72, 33, 27, 36, 106, 17, 141, 39, 125, 41, 37, 25, 47, 61, 376, 107, 31
Offset: 1

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Author

Lee Corbin (lcorbin(AT)rawbw.com), Jan 16 2009

Keywords

Comments

This is presumably a permutation of the positive integers. The inverse permutation (or "index sequence") A322778 begins 4,6,1,10,13,11,2,14,... and gives the position in the continued fraction of Pi at which 1, 2, 3, 4, 5, 6, ... first appear. - Remark corrected by N. J. A. Sloane, Jan 04 2019
The name means that when a number not yet in this sequence appears in the continued fraction of Pi, then that number is added to the sequence. - T. D. Noe, May 06 2013

Examples

			Since the actual continued fraction for Pi is 3, 7, 15, 1, 292, 1, 1, 1, 2, ..., this sequence begins 3, 7, 15, 1, 292, 2, ...
		

Crossrefs

Cf. A001203, A033089 (for records of main continued fraction), A322778 (inverse), A033090.

Programs

  • Mathematica
    DeleteDuplicates[ContinuedFraction[Pi,1000]] (* Harvey P. Dale, May 06 2013 *)
    t = {}; s = ContinuedFraction[Pi, 1000]; Do[If[! MemberQ[t, s[[n]]], AppendTo[t, s[[n]]]], {n, Length[s]}]; t (* T. D. Noe, May 06 2013 *)
  • PARI
    \p 10000
    v=contfrac(Pi); for(i=1,#v,for(j=1,i-1,if(v[i]==v[j],v[i]=0;break))); v=select(n->n,v) \\ Charles R Greathouse IV, May 06 2013

Extensions

More terms from Harvey P. Dale, May 05 2013

A058291 Continued fraction for 2 Pi.

Original entry on oeis.org

6, 3, 1, 1, 7, 2, 146, 3, 6, 1, 1, 2, 7, 5, 5, 1, 4, 1, 2, 42, 5, 31, 1, 1, 1, 6, 2, 2, 4, 3, 12, 49, 1, 5, 1, 12, 1, 1, 1, 2, 3, 1, 2, 1, 1, 3, 1, 16, 2, 1, 1, 15, 2, 3, 6, 3, 8, 18, 6, 1, 2, 1, 3, 2, 1, 2, 2, 1, 1, 2, 1, 8, 2, 33, 1, 80, 91, 1, 10, 1, 5, 1, 2, 2, 2, 1, 49, 2, 8, 2, 3, 5, 4, 2, 1, 1, 1
Offset: 0

Views

Author

Robert G. Wilson v, Dec 07 2000

Keywords

Comments

A001203 is the continued fraction for Pi.

Examples

			6.283185307179586476925286766... = 6 + 1/(3 + 1/(1 + 1/(1 + 1/(7 + ...)))). - _Harry J. Smith_, May 31 2009
		

Crossrefs

Cf. A019692 Decimal expansion. - Harry J. Smith, May 31 2009

Programs

  • Mathematica
    ContinuedFraction[ 2Pi, 100 ]
  • PARI
    contfrac(2*Pi)
    
  • PARI
    { allocatemem(932245000); default(realprecision, 21000); x=contfrac(2*Pi); for (n=0, 20000, write("b058291.txt", n, " ", x[n+1])); } \\ Harry J. Smith, May 31 2009

Extensions

More terms from Jason Earls, Jul 24 2001

A373866 a(n) = maximum element in the continued fraction for Pi truncated to n decimal digits after the decimal point.

Original entry on oeis.org

3, 10, 7, 10, 14, 25, 84, 243, 288, 291, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 351, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292, 292
Offset: 0

Views

Author

Paolo Xausa, Jun 19 2024

Keywords

Examples

			For n = 5, Pi truncated to 5 digits after the decimal point is 3.14159. The corresponding continued fraction is [3, 7, 15, 1, 25, 1, 7], whose maximum element is 25.
		

Crossrefs

Cf. A000796, A001203, A081836 (analogous for phi), A081837 (analogous for e).

Programs

  • Mathematica
    A373866[n_] := Max[ContinuedFraction[Floor[Pi*10^n]/10^n]];
    Array[A373866, 100, 0]

A046956 Numbers k where tan(k) decreases monotonically to 0 (or cot(k) increases).

Original entry on oeis.org

1, 4, 7, 10, 13, 16, 19, 22, 355, 104348, 312689, 1146408, 5419351, 85563208, 165707065, 411557987, 1480524883, 2549491779, 8717442233, 14885392687, 35938735828, 56992078969, 78045422110, 99098765251, 120152108392
Offset: 0

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Author

Keywords

Comments

From Jon E. Schoenfield, Aug 10 2006: (Start)
The approach described uses continued fractions containing an even number of terms of which all but the last term are fixed at the values those terms take in the continued fraction for Pi; the final term is initialized at 1 and incremented by 1 each time until it reaches the value taken by that term in the continued fraction for Pi. The semiconvergents and convergents thus obtained are increasingly accurate approximations for Pi, all of which approach Pi from values larger than Pi. Thus the angles whose sizes (in radians) are the numerators of those semiconvergents and convergents approach (from the positive side) integer multiples of Pi, so the tangents of those angles approach zero from positive values.
If we were to use the same approach but with continued fractions having an odd number of terms, i.e., [3] = 3/1; [3;7,i], i=1..15; [3;7,15,1,i], i=1..292; etc., then the semiconvergents and convergents obtained would likewise be increasingly accurate approximations for Pi, but they would approach Pi from values smaller than Pi, so the angles whose sizes (in radians) are the numerators of those semiconvergents and convergents would approach (from the negative side) integer multiples of Pi and thus the tangents of those angles would approach zero from negative values.
Terms after a(0) = 1 are the numerators of the fractions obtained by evaluating all those convergents and semiconvergents of the continued fraction for Pi (A001203) that, as written below, have an even number of partial quotients:
[3;i], i=1..7 (6 semiconvergents and 1 convergent)
[3;7,15,1]
[3;7,15,1,292,1]
[3;7,15,1,292,1,1,1]
[3;7,15,1,292,1,1,1,2,1]
[3;7,15,1,292,1,1,1,2,1,3,1]
[3;7,15,1,292,1,1,1,2,1,3,1,14,i], i=1..2
[3;7,15,1,292,1,1,1,2,1,3,1,14,2,1,1]
[3;7,15,1,292,1,1,1,2,1,3,1,14,2,1,1,2,i], i=1..2
[3;7,15,1,292,1,1,1,2,1,3,1,14,2,1,1,2,2,2,i], i=1..2
[3;7,15,1,292,1,1,1,2,1,3,1,14,2,1,1,2,2,2,2,1,i], i=1..84, etc. (End)
See also A002485 which has a similar property (numerators of convergents to pi = numbers for which |tan a(n)| decreases to zero). - M. F. Hasler, Apr 01 2013

Examples

			a(1) is the numerator of [3;1] = 3 + 1/1 = 4/1
a(2) is the numerator of [3;2] = 3 + 1/2 = 7/2
...
a(7) is the numerator of [3;7] = 3 + 1/7 = 22/7
a(8) is the numerator of [3;7,15,1] = 3 + 1/(7 + 1/(15 + 1/1)) = 355/113
a(9) is the numerator of [3;7,15,1,292,1] = 3 + 1/(7 + 1/(15 + 1/(1 + 1/(292 + 1/1)))) = 104348/33215
		

Crossrefs

Cf. A001203, A004112, A002485 (|tan a(n)|->0).

Programs

  • Mathematica
    s = Tan[1]; Do[t = Tan[n]; If[t > 0 && t <= s, Print[n]; s = t], {n, 10^9}] (* Ryan Propper, Jul 27 2006 *)
  • PARI
    e=2;for(n=1,1e9, tan(n)>0 && tan(n)M. F. Hasler, Apr 01 2013

Extensions

More terms from Michel ten Voorde
2 more terms from Ryan Propper, Jul 27 2006
More terms from Jon E. Schoenfield, Aug 10 2006
Corrected by Don Reble, Nov 20 2006

A058651 Continued fraction for Pi + e.

Original entry on oeis.org

5, 1, 6, 7, 3, 21, 2, 1, 2, 2, 1, 1, 2, 3, 3, 2, 5, 2, 1, 1, 1, 1, 3, 1, 8, 4, 4, 1, 1, 1, 1, 8, 1, 4, 1, 5, 1, 1, 1, 2, 4, 3, 2, 1, 1, 2, 1, 10, 1, 4, 1, 2, 1, 12, 1, 8, 2, 7, 39, 365, 2, 15, 2, 25, 1, 2, 5, 3, 3, 9, 3, 1, 1, 9, 1, 1, 47, 1, 1, 18, 1, 1, 2, 6, 1, 1, 1, 4, 1, 3, 1, 1, 1, 1, 4, 1, 6, 37
Offset: 0

Views

Author

Avi Peretz (njk(AT)netvision.net.il), Dec 26 2000

Keywords

Comments

The question of the transcendence of the number Pi + e is still open.

Examples

			a(1) = 5 because Pi + e = 5.859874482048838473822930854632165381954416493075065395941912220031...
5.859874482048838473822930854... = 5 + 1/(1 + 1/(6 + 1/(7 + 1/(3 + ...)))). - _Harry J. Smith_, May 31 2009
		

Crossrefs

Cf. A059742 (decimal expansion).

Programs

  • PARI
    \p 500; contfrac(Pi+exp(1))
    
  • PARI
    { allocatemem(932245000); default(realprecision, 21000); x=contfrac(Pi+exp(1)); for (n=1, 20000, write("b058651.txt", n-1, " ", x[n])); } \\ Harry J. Smith, May 31 2009

Extensions

More terms from Jason Earls, Jun 28 2001
Offset changed by Andrew Howroyd, Aug 04 2024

A137299 Square matrix read by antidiagonals: T(m,n) = m-th term in the continued fraction expansion of Pi^n.

Original entry on oeis.org

3, 9, 7, 31, 1, 15, 97, 159, 6, 1, 306, 2, 3, 1, 292, 961, 50, 2, 7, 2, 1, 3020, 2, 1, 3, 1, 47, 1, 9488, 3, 1, 4, 1, 13, 1, 1, 29809, 1, 2, 1, 60, 16539, 2, 8, 2, 93648, 10, 1, 2, 3, 1, 1, 1, 1, 1, 294204, 21, 14, 7, 3, 9, 4, 6, 3, 1, 3, 924269, 55, 15, 1, 1, 2, 1, 23, 7, 1, 2, 1
Offset: 1

Views

Author

M. F. Hasler, Mar 14 2008

Keywords

Comments

The sequence was suggested by Leroy Quet.

Examples

			The matrix limited to order 10 is given by matrix(10,10,m,n,contfrac(Pi^n)[m]):
[   3   9   31    97   306   961  3020  9488 29809 93648]
[   7   1  159     2    50     2     3     1    10    21]
[  15   6    3     2     1     1     2     1    14    15]
[   1   1    7     3     4     1     2     7     1     1]
[ 292   2    1     1    60     3     3     1     9     4]
[   1  47   13 16539     1     9     2     1     3     2]
[   1   1    2     1     4     1    10     3     1     1]
[   1   8    1     6    23     5     4     1     5     3]
[   2   1    3     7     1     1     1     1     8     2]
[   1   1    1     6     2     3     1     1    16     1]
		

Crossrefs

Programs

  • Mathematica
    A137299list[dmax_]:=With[{a=Array[ContinuedFraction[Pi^(dmax+1-#),#]&,dmax]},Array[Diagonal[a,#]&,dmax,1-dmax]];A137299list[10] (* Generates 10 antidiagonals *) (* Paolo Xausa, Nov 14 2023 *)
  • PARI
    concat(vector(20,i,vector(i,j,contfrac(Pi^(i-j+1))[j])))
    
  • PARI
    T(m,n)=contfrac(Pi^n)[m]

A177438 Continued fraction for Pi - sqrt(2).

Original entry on oeis.org

1, 1, 2, 1, 2, 77, 2, 1, 37, 1, 6, 1, 1, 1, 46, 3, 1, 1, 1, 1, 4, 2, 7, 1, 4, 1, 2, 1, 13, 1, 1, 1, 3, 2, 1, 1, 432, 1, 1, 1, 1, 3, 2, 10, 1, 1, 1, 18, 1, 1700, 1, 1, 5, 2, 9, 4, 4, 1, 1, 2, 1, 3, 27, 1, 1, 2, 1, 1, 1, 4, 3, 1, 2, 2, 5, 1, 32, 1, 11, 1, 2, 52, 10, 4, 1, 1, 10, 1, 1, 2, 23, 1, 3, 7, 12, 1
Offset: 0

Views

Author

Earl Bellinger (ebelling(AT)oswego.edu), May 08 2010

Keywords

Crossrefs

Cf. A177437 (decimal expansion of Pi-sqrt(2)), A001203 (continued fraction for Pi), A040000 (continued fraction expansion of sqrt(2)).

Programs

  • Magma
    SetDefaultRealField(RealField(100)); R:= RealField(); ContinuedFraction(Pi(R) - Sqrt(2)); // G. C. Greubel, Sep 29 2018
  • Maple
    with(numtheory): cfrac(Pi-(sqrt(2)),100,'quotients'); # Muniru A Asiru, Sep 29 2018
  • Mathematica
    ContinuedFraction[Pi-Sqrt[2],100] (* Harvey P. Dale, Nov 06 2011 *)
  • PARI
    default(realprecision, 100); contfrac(Pi - sqrt(2)) \\ G. C. Greubel, Sep 29 2018
    

Extensions

Edited and extended by Klaus Brockhaus, May 09 2010

A225802 Position of first occurrence of n in continued fraction for Pi, or -1 if n never occurs.

Original entry on oeis.org

3, 8, 0, 29, 39, 31, 1, 43, 129, 99, 275, 54, 27, 12, 2, 77, 646, 136, 139, 179, 213, 82, 202, 90, 790, 111, 573, 174, 242, 146, 877, 454, 530, 420, 1007, 593, 783, 3040, 720, 1871, 753, 118, 491, 428, 80, 3199, 824, 282, 3026, 464, 1436, 3383, 1546, 1863, 445, 1017
Offset: 1

Views

Author

Eric W. Weisstein, Jul 27 2013

Keywords

Comments

Correctly indexed version of A032523.
All positive integers <= 49003 occur in the first 15000000000 terms of the c.f. (the first that do not are 49004, 50471, 53486, 56315, 58255, ...) - Eric W. Weisstein, Jul 27 2013

Examples

			The continued fraction of Pi is [a_0; a_1, a_2, ...] = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, ...], so
a(1) = 3 (1 first occurs at term a_3);
a(2) = 8 (2 first occurs at term a_8);
a(3) = 0 (3 first occurs at term a_0).
		

Crossrefs

Cf. A032523 (= a(n) + 1).
Cf. A001203 (continued fraction of Pi).

Formula

a(n) = A032523(n) - 1.

Extensions

"Escape clause" added to definition by Jianing Song, Apr 06 2019

A273128 Continued logarithm expansion of Pi.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 3, 0, 3, 0, 2, 0, 0, 2, 5, 0, 1, 2, 0, 1, 2, 4, 3, 0, 0, 1, 2, 0, 0, 1, 7, 3, 2, 3, 0, 1, 0, 0, 0, 0, 3, 2, 3, 1, 3, 2, 0, 0, 0, 0, 2, 2, 0, 0, 0, 1, 0, 1, 2, 1, 0, 3, 4, 0, 0, 0, 3, 0, 0, 0, 1, 1, 1, 2, 2, 0, 3, 0, 0, 1, 0, 2, 1, 3, 0, 0, 2, 0
Offset: 1

Views

Author

Jeffrey Shallit, May 16 2016

Keywords

Crossrefs

Cf. A273127 (similar for e).

Programs

  • PARI
    default(realprecision, 10^5); p=Pi;
    lista(nn) = {my(c); for(n=1, nn, print1(c=logint(p\1, 2), ", "); p=1/(p/2^c-1)); } \\ Jinyuan Wang, Feb 27 2020
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