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 1-3 of 3 results.

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 1-3 of 3 results.

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