A157142 Signed denominators of Leibniz series for Pi/4.
1, -3, 5, -7, 9, -11, 13, -15, 17, -19, 21, -23, 25, -27, 29, -31, 33, -35, 37, -39, 41, -43, 45, -47, 49, -51, 53, -55, 57, -59, 61, -63, 65, -67, 69, -71, 73, -75, 77, -79, 81, -83, 85, -87, 89, -91, 93, -95, 97, -99, 101, -103, 105, -107, 109, -111, 113, -115
Offset: 0
Examples
G.f. = 1 - 3*x + 5*x^2 - 7*x^3 + 9*x^4 - 11*x^5 + 13*x^6 - 15*x^7 + 17*x^8 + ...
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- X. Gourdon and P. Sebah, Archimedes' constant
- Mathpages, How Leibniz might have anticipated Euler
- Wikipedia, Leibniz formula for Pi
- Index entries for linear recurrences with constant coefficients, signature (-2,-1).
Crossrefs
Cf. A157327. [Jaume Oliver Lafont, Mar 03 2009]
Programs
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Magma
[(2*n + 1) * (-1)^n: n in [0..70]]; // Vincenzo Librandi, Dec 23 2018
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Mathematica
a[ n_] := (2*n + 1) * (-1)^n; (* Michael Somos, Nov 21 2022 *)
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PARI
{a(n) = (2*n + 1) * (-1)^n};
Formula
Euler transform of length 2 sequence [-3, 2]. - Michael Somos, Mar 26 2011
a(n) = b(2*n + 1) where b(n) is completely multiplicative with b(2) = 0, b(p) = p if p == 1 (mod 4), b(p) = -p if p == 3 (mod 4). - Michael Somos, Mar 26 2011
With offset 1 this sequence is the exponential reversion of A005264. - Michael Somos, Mar 26 2011
a(-1 - n) = a(n), a(n + 1) + a(n - 1) = -2*a(n) for all n in Z. - Michael Somos, Mar 26 2011
E.g.f.: (1 - 2*x)*exp(-x). - Michael Somos, Mar 26 2011
G.f.: (1 - x)/(1 + x)^2 = (1 - x)^3 / (1 - x^2)^2.
a(0) = 1, a(1) = -3, a(n) = -2*a(n-1) - a(n-2) for n >= 2.
Sum_{n=0..inf} 1/a(n) = Pi/4.
Comments