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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A179617 Continued fraction for Pi^(1/Pi).

Original entry on oeis.org

1, 2, 3, 1, 1, 1, 3, 1, 1, 3, 2, 1, 6, 3, 4, 2, 1, 14, 1, 1, 5, 2, 2, 2, 1, 3, 23, 1, 26, 23, 13, 1, 1, 1, 5, 2, 8, 12, 1, 1, 1, 3, 5, 23, 31, 7, 1, 1, 2, 5, 4, 1, 1, 6, 1, 72, 4, 1, 1, 1, 7, 2, 1, 1, 2, 49, 3, 1, 4, 2, 3, 2, 1, 1, 6, 2, 3, 3, 1, 1, 26, 2, 2, 11, 5, 3, 5, 1, 2, 1, 12, 1, 558, 1, 1, 3, 1, 76
Offset: 0

Views

Author

Vladimir Joseph Stephan Orlovsky, Jul 20 2010

Keywords

Comments

Pi^(1/Pi) = 1.43961949584759068833649080497375567869829...

Crossrefs

Cf. A001203, A073238 (decimal expansion).

Programs

  • Mathematica
    ContinuedFraction[Pi^(1/Pi),200]

Extensions

Offset changed by Andrew Howroyd, Jul 07 2024

A203168 Positions of 1 in the continued fraction expansion of Pi.

Original entry on oeis.org

4, 6, 7, 8, 10, 12, 15, 16, 21, 24, 25, 29, 35, 41, 42, 45, 47, 51, 53, 54, 56, 57, 58, 60, 61, 63, 64, 66, 68, 69, 74, 79, 82, 84, 87, 89, 92, 94, 96, 98, 99, 104, 108, 113, 115, 116, 121, 125, 126, 134, 136, 138, 141, 144, 148, 149, 150, 154, 157, 158, 160
Offset: 1

Views

Author

Ben Branman, Dec 29 2011

Keywords

Comments

In the Gauss-Kuzmin distribution, 1 appears with probability log_2(4/3) = 41.5037...%. Thus the n-th appearance of 1 in the continued fraction of a real number chosen uniformly from [0, 1) will be, with probability 1, n / (log_2(4/3)) + O(sqrt(n)). Does this sequence have the same asymptotic? - Charles R Greathouse IV, Dec 30 2011

Links

Crossrefs

Cf. A001203, A033089, A000796.

Programs

  • Mathematica
    Flatten[Position[ContinuedFraction[Pi, 160], 1]]
  • PARI
    v=contfrac(Pi);for(i=1,#v,if(v[i]==1,print1(i", "))) \\ Charles R Greathouse IV, Dec 30 2011

Formula

A001203(a(n)) = 1.

A236250 Period of the n-th convergent to the continued fraction expansion of Pi.

Original entry on oeis.org

1, 6, 13, 112, 51, 24, 15088, 12284, 88460, 1204, 459, 31824, 93210, 1864254, 531648, 456036, 8299090, 28574910, 1813560, 32552820, 33166008, 133585180, 2503410, 214098720, 3183870690, 7411133309730, 4852769490690, 2294509753536, 175964053944, 3336533898768
Offset: 1

Views

Author

Jani Melik, Jan 21 2014

Keywords

Examples

			The 2nd convergent is 22/7 = 3.142857 142857 ..., whose period is 6, so a(2) = 6.
The 3rd convergent is 333/106 = 3.1 4150943396226 4150943396226 ..., whose period is 13, so a(3) = 13.

Crossrefs

Cf. A001203, A007732, A084407.

Programs

  • Sage
    st_clenov = 30
def A236250(n) :
   vu = continued_fraction_list(pi, nterms=st_clenov);
   p = []
   for i in (0..n) :
      p.append(convergents(vu)[i].period())
   return(p)
A236250(st_clenov-1);

Formula

a(n) = A007732(A002486(n+2)). - Michel Marcus, Jan 21 2014

A282496 'Somos expansion' of Pi: Pi=a(0)*sqrt(a(1)*sqrt(a(2)*sqrt(a(3)*sqrt(...)))). a(n)=floor(x(n)), x(n)=x(n-1)^2/a(n-1)^2, x(0)=Pi.

Original entry on oeis.org

3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 3, 1, 2, 1, 3, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 3, 1, 2, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 2, 2, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1
Offset: 0

Views

Author

Yuriy Sibirmovsky, Feb 16 2017

Keywords

Comments

1<=a(n)<=3 for all n. Reasoning: for x>1 it follows that 1

Examples

			Integer part of Pi is 3. Integer part of Pi^2/9 is 1.

Links

Crossrefs

Cf. A000796 (digits), A100044 (Pi^2/9), A001203 (continued fraction), A276459 (another nested radical expansion).

Programs

  • Mathematica
    $MaxExtraPrecision = 1000;
x00 = Pi;
x0 = x00;
Nm = 130;
j = 1;
Res = Table[1, {j, 1, Nm}];
While[j < Nm, Res[[j]] = Floor[x0]; x0 = N[(x0/ Res[[j]])^2, 20000];
  j++];
Res

Formula

Product_{k>=0} a(k)^(1/2^k) = Pi.

A346534 Denominators of approximations j/k for Pi such that abs(j/k - Pi)*sqrt(5)*k^2 < 1.

Original entry on oeis.org

1, 7, 14, 113, 226, 339, 452, 565, 678, 791, 904, 1017, 1130, 1243, 33215, 99532, 364913, 1725033, 3450066, 25510582, 131002976, 340262731, 811528438, 1963319607, 6701487259, 13402974518, 20104461777, 26805949036, 33507436295, 40208923554, 567663097408
Offset: 1

Views

Author

June Richardson, Jul 22 2021

Keywords

Comments

Define two parameters E and M for a rational approximation j/k for an irrational number x: E = abs(j/k - x) (the absolute error) and M = 1/(sqrt(5)*k^2). Hurwitz's theorem states that every real number has infinitely many rational approximations that satisfy E/M < 1, making each such approximation a "strong approximation". This sequence lists the denominators of such numbers for the irrational number Pi.

Examples

			22/7 ~ 3.1428571 and E/M ~ 0.1385.
355/113 ~ 3.1415929 and E/M ~ 0.0076.
From _Jon E. Schoenfield_, Aug 06 2021: (Start)
    k       j    E = |j/k - Pi|  M = 1/(sqrt(5)*k^2)    E/M
  -----  ------  --------------  -------------------  -------
      1       3  0.141592653590  0.44721359549995794  0.31661
      7      22  0.001264489267  0.00912680807142771  0.13855
     14      44  0.001264489267  0.00228170201785693  0.55419
    113     355  0.000000266764  0.00003502338440755  0.00762
    226     710  0.000000266764  0.00000875584610189  0.03047
    339    1065  0.000000266764  0.00000389148715639  0.06855
    452    1420  0.000000266764  0.00000218896152547  0.12187
    565    1775  0.000000266764  0.00000140093537630  0.19042
    678    2130  0.000000266764  0.00000097287178910  0.27420
    791    2485  0.000000266764  0.00000071476294709  0.37322
    904    2840  0.000000266764  0.00000054724038137  0.48747
   1017    3195  0.000000266764  0.00000043238746182  0.61696
   1130    3550  0.000000266764  0.00000035023384408  0.76167
   1243    3905  0.000000266764  0.00000028944945791  0.92163
  33215  104348  0.000000000332  0.00000000040536522  0.81810
(End)

Links

Crossrefs

Cf. A000796, A001203, A063673.
Cf. A002163 (sqrt(5)).

Programs

  • Magma
    // See Links.
  • Mathematica
    a={}; For[k=1,k<=10^6,k++,If[Abs[Round[k Pi]/k-Pi]Sqrt[5] k^2<1,AppendTo[a,k]]]; a (* Stefano Spezia, Aug 07 2021 *)
  • PARI
    is(k) = my(j=round(Pi*k)); abs(j/k - Pi)*sqrt(5)*k^2 < 1; \\ Jinyuan Wang, Aug 06 2021

Extensions

a(17)-a(19) from Jinyuan Wang, Aug 06 2021
a(20)-a(31) from Jon E. Schoenfield, Aug 06 2021

A365350 Decimal expansion of 1/(Pi-3).

Original entry on oeis.org

7, 0, 6, 2, 5, 1, 3, 3, 0, 5, 9, 3, 1, 0, 4, 5, 7, 6, 9, 7, 9, 3, 0, 0, 5, 1, 5, 2, 5, 7, 0, 5, 5, 8, 0, 4, 2, 7, 3, 4, 3, 1, 0, 0, 2, 5, 1, 4, 5, 5, 3, 1, 3, 3, 3, 9, 9, 8, 3, 1, 6, 8, 7, 3, 5, 5, 5, 9, 0, 3, 3, 3, 7, 5, 8, 0, 0, 5, 6, 0, 8, 3, 5, 0, 3, 9, 7, 7, 4, 7
Offset: 1

Views

Author

Rok Cestnik, Sep 02 2023

Keywords

Comments

The continued fraction expansion is the same as Pi (A001203) with initial 3 omitted.

Examples

			7.062513305931045769793...

Links

Crossrefs

Cf. A000796, A201775, A001203, A194807.

Programs

  • Mathematica
    A365350 = RealDigits[N[1/(Pi-3), #+1]][[1]][[1;; -2]]&;
  • PARI
    1/(Pi-3)

A366397 Decimal expansion of the number whose continued fraction terms are one larger than those of Pi.

Original entry on oeis.org

4, 1, 2, 4, 0, 6, 0, 1, 0, 2, 2, 8, 7, 8, 6, 5, 3, 9, 1, 6, 7, 5, 8, 5, 0, 8, 3, 2, 2, 5, 6, 8, 1, 7, 4, 9, 7, 8, 4, 2, 0, 1, 8, 3, 7, 2, 9, 7, 3, 9, 1, 3, 5, 6, 7, 7, 0, 7, 3, 4, 3, 4, 3, 5, 6, 2, 3, 1, 8, 9, 4, 5, 4, 1, 5, 8, 9, 1, 8, 0, 1, 6, 8, 3, 3, 3, 3, 1, 5, 4, 4, 2, 9, 7, 0, 6, 8, 1, 0, 3, 0, 3, 6, 0
Offset: 1

Views

Author

Rok Cestnik, Oct 08 2023

Keywords

Examples

			4.12406010228786539167585... = 4 + 1/(8 + 1/(16 + 1/(2 + 1/(293 + ...)))).
Pi = 3.141592653589793238... = 3 + 1/(7 + 1/(15 + 1/(1 + 1/(292 + ...)))).

Links

Crossrefs

Cf. A000796, A001203, A059833.

Programs

  • PARI
    N = 25;
cf(v) = my(m=contfracpnqn(v)); m[1, 1]/m[2, 1];
summand(k) = (-1)^k/2^(10*k)*(-2^5/(4*k+1)-1/(4*k+3)+2^8/(10*k+1)-2^6/(10*k+3)-2^2/(10*k+5)-2^2/(10*k+7)+1/(10*k+9));
pi1 = contfrac(1/2^6*sum(k=0,N,summand(k)));
pi2 = contfrac(1/2^6*sum(k=0,N+1,summand(k)));
n = 0; while(pi1[1..n+1] == pi2[1..n+1], n++);
ap1 = cf(apply(x->x+1, pi1[1..n-1]));
ap2 = cf(apply(x->x+1, pi1[1..n]));
n = 0; while(digits(floor(10^(n+1)*ap1)) == digits(floor(10^(n+1)*ap2)), n++);
A366397 = digits(floor(10^n*ap1));

A053431 Numbers k such that (Pi/2)*k^2*sin(1/k) < floor(Pi*k/2).

Original entry on oeis.org

78256779, 340262731, 1963319607, 13402974518, 26805949036, 40208923554, 5703436923116, 136308121570117, 272616243140234, 1684937174853026414, 3369874349706052828, 5054811524559079242, 75474011359728834791267, 909474452321624805685313, 1818948904643249611370626
Offset: 1

Views

Author

N. J. A. Sloane, Mar 21 2000

Keywords

Comments

The solution uses sequence A001203, the continued fraction expansion of Pi. - T. D. Noe, Mar 13 2014

Links

Crossrefs

Cf. A001203.

Extensions

More terms from Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), May 25 2000
Better name from T. D. Noe, Mar 13 2014
More terms from Sean A. Irvine, Dec 23 2021

A071940 Number of 1's among the first n terms of the simple continued fraction for Pi.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 3, 4, 4, 5, 5, 6, 6, 6, 7, 8, 8, 8, 8, 8, 9, 9, 9, 10, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 15, 15, 15, 16, 16, 17, 17, 17, 17, 18, 18, 19, 20, 20, 21, 22, 23, 23, 24, 25, 25, 26, 27, 27, 28, 28, 29, 30, 30, 30, 30, 30, 31, 31, 31, 31
Offset: 1

Views

Author

Benoit Cloitre, Jun 15 2002

Keywords

Examples

			The continued fraction for Pi begins: 3, 7, 15, 1, 292, 1, 1, ...; there are 3 "1's" among the first 7 terms, hence a(7)=3.

Crossrefs

Cf. A001203.

Programs

  • Mathematica
    Accumulate[If[#>1,0,1]&/@ContinuedFraction[Pi,100]] (* Harvey P. Dale, Feb 27 2023 *)
  • PARI
    for(n=1,100,print1(sum(i=1,n,if(component(contfrac(Pi),i)-1,0,1)),","))

A091658 When A032523 is a maximum; or, A091657 less duplicates.

Original entry on oeis.org

4, 9, 30, 40, 44, 130, 276, 647, 791, 878, 1008, 3041, 3200, 3384, 5606, 9721, 17899, 22640, 34070, 34152, 37648, 91193, 134943, 152617, 158172, 190950, 258992, 315679, 525765, 558041, 734305, 1500708, 1669873, 1873804, 1936902, 4278672, 5227319, 7385934, 7876549, 10765774, 11396841, 11466234, 12994613, 19147251, 31403937, 43166470
Offset: 1

Views

Author

Robert G. Wilson v, Jan 26 2004

Keywords

Comments

Each entry is enumerated: 1,2,1,2,1,1,2,6,8,4,1,1,1,1,1,1,1,1,1,8,6,... in A091657.
The 4278672nd term of the continued fraction expansion of Pi is 837.

Examples

			One has to go to the 30th term of the continued fraction of Pi (4) to have seen the integers 1, 2, 3 & 4.

Links

Crossrefs

Cf. A001203, A032523, A091657.

Programs

  • Mathematica
    cfpi = ContinuedFraction[Pi, 10000000]; a = Table[0, {1562}]; Do[b = cfpi[[n]]; If[b < 1563 && a[[b]] == 0, a[[b]] = n], {n, 1, 10000000}]; c
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Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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