A144874 Coefficients of the series expansion of q^(-1/4) pi_q.
1, 2, 0, 0, 1, -2, 1, 2, -3, 0, 2, 0, -1, 0, -1, 0, 4, -2, -2, 0, -1, 4, 1, -4, 0, 2, -2, 0, 2, 0, -1, 2, -1, -4, 2, 0, 2, 2, -2, 0, -2, -2, 3, 2, -3, 0, 4, -2, -2, 2, -2, 2, 0, -4, 0, 4, 3, -2, -1, -2, 0, 2, -2, -2, 2, 2, 2, 0, -4, 0, 2, -2, 1, 2, -3, -2, 4, 0, -2, 2, -2, 4, 0, -4, 2, -2, -2, 2, 2, -2, -1, 4, 1, -2, 2, -2, -4, 2, 0, 0, 2
Offset: 0
Keywords
Examples
G.f. = 1 + 2*x + x^4 - 2*x^5 + x^6 + 2*x^7 - 3*x^8 + 2*x^10 + ...
References
- R. Roy, Sources in the development of mathematics, Cambridge University Press 2011.
- R. W. Gosper, Experiments and discoveries in q-trigonometry, in Symbolic Computation, Number Theory, Special Functions, Physics and Combinatorics. Editors: F. G. Garvan and M. E. H. Ismail. Kluwer, Dordrecht, Netherlands, 2001, pp. 79-105.
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..10000
- R. W. Gosper, Experiments and discoveries in q-trigonometry, Preprint.
- R. W. Gosper, q-Trigonometry: Some Prefatory Afterthoughts
- K. Ono, S. Robins and P. T. Wahl, On the representation of integers as sums of triangular numbers, aequationes mathematicae, August 1995, Volume 50, Issue 1-2, pp 73-94
- Eric Weisstein's World of Mathematics, q-Pi
- Wikipedia, Leibniz formula for Pi
- Wikipedia, q-analog
- Wikipedia, q-gamma function
- Wikipedia, Wallis product
Crossrefs
Cf. A008441.
Programs
-
Mathematica
max = 100; pi[q_] := (1 - q^2)*q^(1/4)*Product[(1 - q^(2n))^2 / (1 - q^(2n - 1))^2, {n, 1, max}]; CoefficientList[ Series[ q^(-1/4)*pi[q], {q, 0, max}], q] (* Jean-François Alcover, Feb 07 2013 *)
Formula
From Peter Bala, Dec 12 2013: (Start)
Pi(q) = q^(1/4)*pi_q.
Pi(q) = (1 - q^2)*( Sum_{n >=0} q^(n*(n+1)/2) )^2.
Some q-analogs of classical formulas
= = = = = = = = = = = = = = = = = = =
Let [n] := 1 + q + q^2 + ... + q^(n-1) denote the q-analog of the natural number n.
(a) Wallis' formula Pi/2 = (2/1)*(2/3)*(4/3)*(4/5)*(6/5)*(6/7)* ....
q_analog: Pi(q)/[2] = ([2]/[1])*([2]/[3])*([4]/[3])*([4]/[5])*([6]/[5])*([6]/[7])* ....
(b) The Euler-Sylvester continued fraction Pi/2 = 1 + 1/(1 + 2/(1 + 6/(1 + 12/(1 + ...)))) (Roy 3.47 and 3.67).
q-analog: Pi(q)/[2] = 1 + q/(1 + q*[1]*[2]/(1 + q*[2]*[3]/(1 + q*[3]*[4]/(1 + ...)))).
(c) The Madhava-Leibniz series Pi/4 = 1 - 1/3 + 1/5 - 1/7 + ....
We have two q-analogs:
Pi(q^2)/[4] = 1/[1] - q/[3] + q^2/[5] - q^3/[7] + ...,
as well as
Pi(q)/[2] = sum {n in Z} (-1)^n*q^(n*(n+1))/[2*n+1].
(d) The result Pi^2/8 = sum {n >= 0} 1/(2*n+1)^2.
q-analog: Pi(q^2)^2/[2]^2 = (1 + q)/[1]^2 + q*(1 + q^3)/[3]^2 + q^2*(1 + q^5)/[5]^2 + ....
(e) The result Pi^4/96 = sum {n >= 0} 1/(2*n+1)^4.
q-analog: q*Pi(q^2)^4/[2]^4 = f(q)/[1]^4 + f(q^3)/[3]^4 + f(q^5)/[5]^4 + ..., where f(q) = q + 4*q^2 + q^3. (End)
Comments