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 1-4 of 4 results.

A093828 Decimal expansion of (3*Pi)/8.

Original entry on oeis.org

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

Views

Author

Eric W. Weisstein, Apr 16 2004

Keywords

Comments

Area of an astroid with a = 1.

Examples

			1.1780972450961724644234912687298135815739385247656646...

Links

Crossrefs

Cf. A161685 (continued fraction). - Harry J. Smith, Jun 18 2009

Programs

  • Magma
    SetDefaultRealField(RealField(110)); R:= RealField(); 3*Pi(R)/8; // G. C. Greubel, Aug 11 2019
  • Maple
    evalf[110](3*Pi*(1/8)); # G. C. Greubel, Aug 11 2019
  • Mathematica
    RealDigits[3*Pi/8, 10, 105][[1]] (* G. C. Greubel, Aug 11 2019 *)
  • PARI
    { default(realprecision, 20080); x=3*Pi/8; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b093828.txt", n, " ", d)); } \\ Harry J. Smith, Jun 18 2009
  • Sage
    numerical_approx(3*pi/8, digits=110) # G. C. Greubel, Aug 11 2019

Formula

Equals Integral_{x>0} sin(x)^3/x^3. - Jean-François Alcover, Jun 04 2013
From Amiram Eldar, Aug 02 2020: (Start)
Equals arctan(1 + sqrt(2)).
Equals Integral_{x=0..1} x^(3/2)/sqrt(1-x) dx. (End)
Equals Sum_{k>=1} sin(k*Pi/4)/k. - Amiram Eldar, May 30 2021
3*Pi/8 = Sum_{n >= 1} n*(n+1)*2^(n+1)/binomial(2*n+6,n+3) (apply Euler's series transformation to the series representation Pi = 384*Sum_{n >= 1} (-1)^(n+1)*n^2/((4*n^2 - 1)*(4*n^2 - 9)*(4*n^2 - 25)) ). - Peter Bala, Dec 08 2021

A195697 First denominator and then numerator in a fraction expansion of log(2) - Pi/8.

Original entry on oeis.org

2, 1, 3, -1, 12, 1, 30, 1, 35, -1, 56, 1, 90, 1, 99, -1, 132, 1, 182, 1, 195, -1, 240, 1, 306, 1, 323, -1, 380, 1, 462, 1, 483, -1, 552, 1, 650, 1, 675, -1, 756, 1, 870, 1, 899, -1, 992, 1, 1122, 1, 1155, -1, 1260
Offset: 1

Views

Author

Mohammad K. Azarian, Sep 25 2011

Keywords

Comments

The minus sign in front of a fraction is considered the sign of the numerator.

Examples

			1/2 - 1/3 + 1/12 + 1/30 - 1/35 + 1/56 + 1/90 - 1/99 + 1/132 + 1/182 - 1/195 + 1/240 + ... = [(1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + (1/9 - 1/10) + (1/11 - 1/12) + ... ] - (1/2)*[(1 - 1/3) + (1/5 - 1/7) + (1/9 - 1/11) + (1/13 - 1/15) + ... ] = log(2) - Pi/8.

References

  • Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
  • Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968).

Crossrefs

Cf. A164833, A195908, A195909, A195913, A016655, A019675, A075549, A098289, A118324, A144981, A161685, A168056, A004772.

Formula

log(2) - Pi/8 = Sum_{n>=1} (-1)^(n+1)*(1/n) + (-1/2)*Sum_{n>=0} (-1)^n*(1/(2*n+1)).
Empirical g.f.: x*(2+x+x^2-2*x^3+9*x^4+2*x^5+14*x^6-2*x^7+3*x^8+2*x^9+3*x^10-2*x^11+x^13) / ((1-x)^3*(1+x)^3*(1-x+x^2)^2*(1+x+x^2)^2). - Colin Barker, Dec 17 2015

A195909 First numerator and then denominator in a fraction expansion of log(2) - Pi/8.

Original entry on oeis.org

1, 2, -1, 3, 1, 12, 1, 30, -1, 35, 1, 56, 1, 90, -1, 99, 1, 132, 1, 182, -1, 195, 1, 240, 1, 306, -1, 323, 1, 380, 1, 462, -1, 483, 1, 552, 1, 650, -1, 675, 1, 756, 1, 870, -1, 899, 1, 992, 1, 1122, -1, 1155, 1
Offset: 1

Views

Author

Mohammad K. Azarian, Sep 26 2011

Keywords

Examples

			1/2 - 1/3 + 1/12 + 1/30 - 1/35 + 1/56 + 1/90 - 1/99 + 1/132 + 1/182 - 1/195 + 1/240 + ... = [(1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + (1/9 - 1/10) + (1/11 - 1/12) + ... ] - (1/2)*[(1 - 1/3) + (1/5 - 1/7) + (1/9 - 1/11) + (1/13 - 1/15) + ... ] = log(2) - Pi/8.

References

  • Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
  • Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968).

Crossrefs

Cf. A195913, A195697, A195947, A164833, A118324, A098289, A075549, A016655, A019675, A161685, A144981, A168056, A004772.

Formula

log(2) - Pi/8 = Sum_{n>=1} (-1)^(n+1)*(1/n) + (-1/2)*Sum_{n>=0} (-1)^n*(1/(2*n+1)).
Empirical g.f.: x*(1+2*x-2*x^2+x^3+2*x^4+9*x^5-2*x^6+14*x^7+2*x^8+3*x^9-2*x^10+3*x^11+x^12) / ((1-x)^3*(1+x)^3*(1-x+x^2)^2*(1+x+x^2)^2). - Colin Barker, Dec 17 2015

A195913 The denominator in a fraction expansion of log(2)-Pi/8.

Original entry on oeis.org

2, 3, 12, 30, 35, 56, 90, 99, 132, 182, 195, 240, 306, 323, 380, 462, 483, 552, 650, 675, 756, 870, 899, 992, 1122, 1155, 1260, 1406, 1443, 1560, 1722, 1763, 1892, 2070, 2115, 2256, 2450, 2499, 2652, 2862, 2915
Offset: 1

Views

Author

Mohammad K. Azarian, Sep 25 2011

Keywords

Comments

The minus sign in front of a fraction is considered the sign of the numerator and hence the sign of the fraction does not appear in this sequence.

Examples

			1/2 - 1/3 + 1/12 + 1/30 - 1/35 + 1/56 + 1/90 - 1/99 + 1/132 + 1/182 - 1/195 + 1/240 + ... = [(1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + (1/9 - 1/10) + (1/11 - 1/12) + ...] - (1/2)*[(1 - 1/3) + (1/5 - 1/7) + (1/9 - 1/11) + (1/13 - 1/15) + ... ] = log(2) - Pi/8.

References

  • Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968).

Links

  • Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.

Crossrefs

Cf. A195909, A195697, A195947, A164833, A118324, A098289, A075549, A016655, A019675, A161685, A144981, A168056, A004772.

Formula

log(2) - Pi/8 = Sum_{n>=1} (-1)^(n+1)*(1/n) + (-1/2)*Sum_{n>=0} (-1)^n*(1/(2*n+1)).
Empirical g.f.: x*(2+x+9*x^2+14*x^3+3*x^4+3*x^5) / ((1-x)^3*(1+x+x^2)^2). - Colin Barker, Dec 17 2015
From Bernard Schott, Aug 11 2019: (Start)
k >= 1, a(3*k) = (4*k-1) * 4*k,
k >= 0, a(3*k+1) = (4*k+1) * (4*k+2),
k >= 0, a(3*k+2) = (4*k+1) * (4*k+3).
The even terms a(3*k) and a(3*k+1) come from log(2) and the odd terms a(3*k+2) come from - Pi/8. (End)
Showing 1-4 of 4 results.

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