A duality correspondence is set up between polynomials satisfying a recurrence relation and the generalized eigenvectors of a Jacobi operator in a rigged Hilbert space.
Sundt and Jewell (1981) investigated a family of distributions satisfying probability recurrence relation.
Lemma 2 Let m ≥ 2 and let { u n } n ≥ 0 be an integer sequence satisfying the recurrence formula (1).
Notice that for Divide and Conquer algorithms with running time satisfying the recurrence equation (1), it is typically sufficient to obtain the complexity on inputs of size n, where n ranges over the set N b [45, 46].
If there exists a sequence { c m } satisfying the recurrence relation ( m + 2 ) ( m + 1 ) c m + 2 + ∑ k = 0 m [ ( k + 1 ) c k + 1 q m − k + c k r m − k ] = a m (5).
Specifically, suppose that ∥ x ∥ = ⌊ x + 1 2 ⌋ (the nearest integer function), and { v n } n ≥ 0 is an integer sequence satisfying the recurrence formula v n = a 1 v n − 1 + a 2 v n − 2 + ⋯ + a s v n − s ( s ≥ 2 ).
Theorem 2.1 Assume that the radius of convergence of power series ∑ m = 0 ∞ a m x m is ρ 1 > 0 and that there exists a sequence { c m } satisfying the recurrence relation ∑ k = 0 m [ ( k + 2 ) ( k + 1 ) c k + 2 p m − k + ( k + 1 ) c k + 1 q m − k + c k r m − k ] = a m (3).
Therefore Corollary 10, and with it Theorem 9, retrieve as a particular caseTheorem 6 given in [4], which was obtained with the aim of analyzing the asymptotic complexity ofthose algorithms with running time of computing satisfying the recurrence equation (2) bymeans of fixed-point arguments based on the Baire partial quasi-metric space.
So f w can be identified with the running time of computing of arecursive algorithm satisfying the recurrence equation (8), say f T. Of course, f T ( n ) = f w ( n ) = w n for all n ∈ N. Next assume that there exists g ∈ RT c, k such that Γ T ( g ) ≤ g.
Assume that the radius of convergence of power series ∑ m = 0 ∞ a m x m is ρ 1 > 0 and that there exists a sequence { c m } satisfying the recurrence formula ∑ k = m 0 m [ ( k + 2 ) ( k + 1 ) c k + 2 p m − k + ( k + 1 ) c k + 1 q m − k + c k r m − k ] = a m (4).
That is, they studied the computational problem of the nearest integer function of ( ∑ k = n ∞ 1 u k ) − 1 and proved an interesting conclusion: ∥ ( ∑ k = n ∞ 1 u k ) − 1 ∥ = u n − u n − 1 for all n > n 0, where ∥ ⋅ ∥ denotes the nearest integer, namely ∥ x ∥ = ⌊ x + 1 2 ⌋, { u n } n ≥ 0 is an integer sequence satisfying the recurrence formula u n = a u n − 1 + u n − 2 + ⋯ + u n − s ( s ≥ 2 ).
