Abstract
If fn is a free parameter, we give a combinatorial closed form solution of the recursion (n + 1)2un+1 − fnun − n 2un−1 = 0, n ≥ 1, and a related generating function. This is used to give a solution to the Apery type sequence rnn3 + rn−1nαn3 −3α2n+α+ 2θon − θo+ rn−2(n − 1)3 = 0, n ≥ 2, for certain parameters α, θ. We show from another viewpoint two independent solutions of the last recursion related to certain modular forms associated with a problem of conformal mapping: Let f(τ) be a conformal map of a zero-angle hyperbolic quadrangle to an open half plane with values 0, ρ, 1, ∞ (0 < ρ < 1) at the cusps and define t = t(τ) := 1ρf(τ)f(τ)−ρ(τ)−1. Then the function E(τ) = 12πif0(τ)f(τ)11 −f(τ)ρ is a solution, as a generating function in the variable t, of the above recurrence. In other words, E(τ) = r0 +r1t+r2t 2 +. . . , where r0 = 1, r1 = −θ, α = 2− 4ρ.
Highlights
Let P (n) be the third degree polynomial in n defined byP (n) = αn3 + 3α n2 + α + 2θ n + θ, (1)with α, θ complex or real numbers
One should notice that P (n − 1) = −P (−n) = αn3 − 3α n2 + 2 α + 2θ n − θ
This paper is devoted to the study of sequences (r) = (r0, r1, r2, . . .) defined by rnn3 + rn−1P (n − 1) + rn−2(n − 1)3 = 0, n ≥ 2
Summary
With α, θ complex or real numbers. One should notice that P (n − 1) = −P (−n) = αn3 − 3α n2 + 2 α + 2θ n − θ. This paper is, in some sense, an attempt to find the solutions (a), (b) of the recurrence (2) in closed form. In Theorem 1, which we believe is interesting in its own right, we solve the easier recursion (n + 1)2un+1 − fnun − n2un−1 = 0, where fn is a free parameter This result can be seen as a variant of a certain recursion given in an interesting paper of A. This solves a particular case of Heun’s equation. As shown in Theorem 3, we construct the solutions of the recursion (2) as a generating function of certain modular forms attached to f (τ ) with certain parameters α, θ depending on this last function
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