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)

Links

Crossrefs

Cf. A001466, A156269, A156618.
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

A156750 Greedy Egyptian fraction for Pi/2.

Original entry on oeis.org

1, 2, 15, 243, 69282, 36600664305, 6435072487994269232829, 364103502021384610224777078613738432189483892
Offset: 0

Views

Author

Jaume Oliver Lafont, Feb 14 2009

Keywords

Links

Crossrefs

Cf. A001466, A156269, A156736.

Programs

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

Formula

Sum(n>=0,1/a(n))=Pi/2

A164916 Denominators of a BBP series for Pi/4.

Original entry on oeis.org

1, -8, -20, -24, 144, -384, -832, -896, 4352, -10240, -21504, -22528, 102400, -229376, -475136, -491520, 2162688, -4718592, -9699328, -9961472, 42991616, -92274688, -188743680, -192937984, 822083584, -1744830464, -3556769792
Offset: 0

Views

Author

Jaume Oliver Lafont, Aug 31 2009

Keywords

Comments

From the BBP formula for Pi, the following expression for Pi/4 in unit numerators is obtained
Pi/4 = Sum((1/(8k+1)+1/(-2*(8k+4))+1/(-4*(8k+5))+1/(-4*(8k+6)))/16^k, k>=0)
Therefore a(n) such that
a(4*n) = (8*n+1)*16^n.
a(4*n+1) = -2*(8*n+4)*16^n.
a(4*n+2) = -4*(8*n+5)*16^n.
a(4*n+3) = -4*(8*n+6)*16^n.
has
Sum_{n >= 0} (1/a(n)) = Pi/4.
Using PARI/GP suminf(n=0,1/(2^(n-2)*(2*(-1+(-1)^n+(1-I)*(-I)^n+(1+I)*I^n)+(-3+3*(-1)^n+(4-I)*(-I)^n+(4+I)*I^n)*n)))= 0.7853981633974483096156608454...=Pi/4. - Alexander R. Povolotsky, Sep 01 2009

Links

Crossrefs

Cf. A048581, A066968, A154925, A154962, A156269, A157142.

Programs

  • Mathematica
    CoefficientList[Series[(1 - 8*x - 20*x^2 - 24*x^3 + 112*x^4 - 128*x^5 - 192*x^6 - 128*x^7)/(1 - 16*x^4)^2, {x,0,50}], x] (* G. C. Greubel, Feb 25 2017 *)
  • PARI
    x='x + O('x^50); Vec((1 - 8*x - 20*x^2 - 24*x^3 + 112*x^4 - 128*x^5 - 192*x^6 - 128*x^7)/(1 - 16*x^4)^2) \\ G. C. Greubel, Feb 25 2017

Formula

G.f.: (1-8*x-20*x^2-24*x^3+112*x^4-128*x^5-192*x^6-128*x^7)/(1-16*x^4)^2.
a(n)= 2^(n-2)*(2*(-1+(-1)^n+(1-I)*(-I)^n+(1+I)*I^n)+(-3+3*(-1)^n+(4-I)*(-I)^n+(4+I)*I^n)*n). - Alexander R. Povolotsky, Sep 01 2009

Extensions

Comment section corrected by Jaume Oliver Lafont, Sep 03 2009
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.