A154962 -id:A154962 - 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)

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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