Abstract

Abstract. For the solution {u n } ∞n=0 to the polynomial recursion (n + 1) 5 u n+1 −3(2n + 1)(3n 2 + 3n + 1)(15n 2 + 15n + 4)u n − 3n 3 (3n − 1)(3n + 1)u n−1 = 0, wheren = 1,2,..., with the initial data u 0 = 1, u 1 = 12, we prove that all u n are integers.The numbers u n , n = 0,1,2,..., are denominators of rational approximations to ζ(4)(see math.NT/0201024). We use Andrews’s generalization of Whipple’s transformationof a terminating 7 F 6 (1)-series and the method from math.NT/0311114. Consider the following 3-term polynomial recursion:(n+ 1) 5 u n+1 − 3(2n+ 1)(3n 2 + 3n+1)(15n 2 + 15n+ 4)u n −3n 3 (3n− 1)(3n+1)u n−1 = 0 for n > 1,and take the two linearly independent solutions {u n } ∞n=0 and {v n } ∞n=0 determinedby the inicial conditions u 0 = 1, u 1 = 12 and v 0 = 0, v 1 = 13. In [Z1], we give ahypergeometric interpretation of the sequence u n ζ(4)−v n , n = 0,1,2,..., from whichone obtains the limitlim n→∞ v n u n = ζ(4) =π 4 90and the representationu n = (−1) n+1 X nl=0 ddln2−lnl

Full Text
Paper version not known

Talk to us

Join us for a 30 min session where you can share your feedback and ask us any queries you have

Schedule a call

Disclaimer: All third-party content on this website/platform is and will remain the property of their respective owners and is provided on "as is" basis without any warranties, express or implied. Use of third-party content does not indicate any affiliation, sponsorship with or endorsement by them. Any references to third-party content is to identify the corresponding services and shall be considered fair use under The CopyrightLaw.