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 1-2 of 2 results.

A048581 Numerators of b(n) = (1/16^n)*(4/(8*n+1) - 2/(8*n+4) - 1/(8*n+5) - 1/(8*n+6)).

Original entry on oeis.org

47, 53, 829, 79, 857, 1901, 5273, 97, 1787, 5563, 4519, 4057, 19139, 743, 25681, 229, 3687, 18647, 8329, 3853, 51067, 28069, 20483, 335, 72791, 4379, 85093, 22901, 6557, 52673, 112577, 2501, 127759, 13571, 15989, 38083, 161003, 28319, 35813
Offset: 0

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Author

Benoit Cloitre, Aug 13 2002

Keywords

Comments

Sum_{k>=0} b(k) = Pi was the first BBP formula for Pi (Bayley-Borwein-Plouffe in 1995). Allows one to extract any specified binary digit of Pi.

Links

Crossrefs

Cf. A066968.

Programs

  • Mathematica
    Numerator[Table[1/16^n*(4/(8*n + 1) - 2/(8*n + 4) - 1/(8*n + 5) - 1/(8*n + 6)), {n, 0, 100}]] (* G. C. Greubel, Feb 18 2017 *)
  • PARI
    a(n)=numerator(1/16^n*(4/(8*n+1)-2/(8*n+4)-1/(8*n+5)-1/(8*n+6)))
  • PARI
    a(n)=numerator((1/16)^n*sum(i=1,4,((-1)^(ceil(4/(2*i))))*(floor(4/i))/(8*n+i+floor(sqrt(i-1))*(floor(sqrt(i-1))+1)))) \\ Alexander R. Povolotsky, Aug 31 2009

Formula

Sum_{k>=0} b(k) = Pi.
a(n) = numerator((1/16)^n*sum(i=1,4,((-1)^(ceiling(4/(2*i))))*(floor(4/i))/(8*n+i+floor(sqrt(i-1))*(floor(sqrt(i-1))+1)))). - Alexander R. Povolotsky, Aug 31 2009

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 1-2 of 2 results.

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