A154925 -id:A154925 - cp's OEIS frontend

A156269 Denominators of a series expansion for Pi/2.

1, 2, 6, -20, -24, -56, 144, 160, 352, -832, -896, -1920, 4352, 4608, 9728, -21504, -22528, -47104, 102400, 106496, 221184, -475136, -491520, -1015808, 2162688, 2228224, 4587520, -9699328, -9961472, -20447232, 42991616, 44040192, 90177536

Offset: 0

Jaume Oliver Lafont, Feb 07 2009, Feb 21 2009

Numerators are all 1.
Sum_{n >= 0} 1/a(n) = Pi/2.
This sequence is based on Adamchik and Wagon's BBP-type three-term formula for Pi, namely Pi = Sum_{n >= 0} (-1/4)^n*(2/(4*n + 1) + 2/(4*n + 2) + 1/(4*n + 3)).
From Peter Bala, Jun 16 2016: (Start)
The reciprocals 1/a(n) appear as coefficients in the Maclaurin series for 2*arctan(z/(2 - z)) = z + z^2/2 + z^3/6 - z^5/20 - z^6/24 - z^7/56 + ... (the radius of convergence is sqrt(2)).
Setting z = 1 gives Pi/2 = Sum_{n >= 0} 1/a(n) as observed above. Setting z = 2 - sqrt(2) gives a series for Pi/4 in terms of a(n). Setting z = +- sqrt(2), and using Abel's theorem on power series, gives two further series for Pi involving a(n). (End)

V. Adamchik and S. Wagon, Pi: A 2000-Year Search Changes Direction
Index entries for linear recurrences with constant coefficients, signature (0, 0, -8, 0, 0, -16).

Cf. A000796, A154925, A154962.

A156269 := n -> if `mod`(n, 3) = 0 then (-4)^(n/3)*(4*n/3 + 1) elif `mod`(n, 3) = 1 then (-4)^((n-1)/3)*(4*(n-1)/3 + 2) else (-4)^((n-2)/3)*(8*(n-2)/3 + 6) end if:
seq(A156269(n), n = 1 .. 35); # Peter Bala, Jun 16 2016

CoefficientList[Series[(1+2x+6x^2-12x^3-8x^4-8x^5)/(1+4x^3)^2,{x,0,40}],x] (* or *) LinearRecurrence[{0,0,-8,0,0,-16},{1,2,6,-20,-24,-56},40] (* Harvey P. Dale, Dec 16 2016 *)

G.f.: (1+2*x+6*x^2-12*x^3-8*x^4-8*x^5)/(1+4*x^3)^2.
From Peter Bala, Jun 16 2016: (Start)
a(3*n) = (-4)^n*(4*n + 1);
a(3*n + 1) = (-4)^n*(4*n + 2);
a(3*n + 2) = (-4)^n*(8*n + 6). (End)

A154962 The terms of this sequence are integer values of consecutive denominators (with signs) from the fractional expansion (using only fractions with numerators to be positive 1's) of the BBP polynomial ( 4/(8k+1) - 2/(8k+4) - 1/(8k+5) - 1/(8k+6) ) for all k (starting from 0 to infinity); for k>=1 the Erdos-Straus conjecture is applied to the first fraction - so it is always replaced by exactly three fractions.

Original entry on oeis.org

1, 1, 1, 1, -2, -5, -6, 3, 10, 90, -5, -13, -14, 5, 30, 510, -10, -21, -22, 7, 60, 2100, -14, -29, -30

Offset: 0

Alexander R. Povolotsky, Jan 18 2009, corrected Jan 20 2009

sign

This sequence is different from A154925, where the first fraction for k>=1 is expanded with Egyptians fractions, using R.Knott's converter calculator #1 (http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html#calc1)

Cf. A073101, A075245, A075246, A075247, A154925

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

Jaume Oliver Lafont, Aug 31 2009

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

G. C. Greubel, Table of n, a(n) for n = 0..1000

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

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 *)

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

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

Comment section corrected by Jaume Oliver Lafont, Sep 03 2009

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A156269 Denominators of a series expansion for Pi/2.

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A164916 Denominators of a BBP series for Pi/4.

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