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.

A156736 Signed greedy Egyptian fraction for Pi/2.

Original entry on oeis.org

1, 2, 14, -1582, -7497258, 303297921775458, -2646995089135122277190614296178, 82888930564911423983289917045230098319343306166666586941750246
Offset: 0

Views

Author

Jaume Oliver Lafont, Feb 14 2009

Keywords

Comments

The second and fourth convergents of Pi (22/7 and 355/113) appear when truncating the series to three and four terms.

Examples

			1+1/2+1/14=11/7=(1/2)(22/7)
1+1/2+1/14-1/1582=355/226=(1/2)(355/113)
		

Crossrefs

Cf. A156750. [From Jaume Oliver Lafont, Mar 03 2009]

Programs

  • PARI
    x=Pi/2; for (k=0,7, d=round(1/x); x=x-1/d; print1(d,", "))

Formula

Sum(n>=0,1/a(n))=Pi/2.
a(n) = 2*A001467(n+1). - R. J. Mathar, Apr 02 2011

A157193 Greedy Egyptian fraction expansion for 4/Pi.

Original entry on oeis.org

1, 4, 44, 1953, 4179942, 42836179578838, 3958573977160882295479936105, 36328295343356352083453782833218820307659379901717630389
Offset: 0

Views

Author

Jaume Oliver Lafont, Feb 24 2009

Keywords

Comments

Sum_{n>=0}1/a(n)=4/Pi
Truncating the series to three terms yields the convergent 22/7 as an approximation to Pi:
1+1/4+1/44=14/11=4/(22/7)

Crossrefs

Programs

  • PARI
    x=4/Pi; for (k=0,7,d=ceil(1/x);x=x-1/d;print(d,", "))

A164924 Denominators of Egyptian fraction for Pi/4 whose partial sums are the convergents.

Original entry on oeis.org

1, -4, 20, -45, 126, -3066, 98988, -14808876, 1088223045, -11017349070, 121040912274, -2384786370947, 45093873748908, -140317516983756, 553849385827045, -4262743007968170, 28759236845787534, -62522995423971463
Offset: 0

Views

Author

Jaume Oliver Lafont, Aug 31 2009

Keywords

Comments

1-1/4+1/20-1/45+1/126 = (22/7)/4
1-1/4+1/20-1/45+1/126-1/3066+1/98988 = (355/113)/4

Crossrefs

Cf. A156618.

Programs

  • PARI
    c0=0; for (k=2, 30, m=contfracpnqn(contfrac(Pi/4, k)); c1=m[1, 1]/m[2, 1]; print1(1/(c1-c0), ", "); c0=c1; )
Showing 1-3 of 3 results.