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

Author

Jaume Oliver Lafont, Feb 07 2009, Feb 21 2009

Keywords

• sign
• easy

Comments

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)

Links

• 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).

Crossrefs

Cf. A000796, A154925, A154962.

Programs

Maple

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

Mathematica

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 *)

Formula

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)