Sentence examples for recurrence sequences from inspiring English sources

We consider the following type of higher-order recurrence sequences.

Recently, many authors have studied some special linear recurrence sequences in algebraic structures; see, for example, [4 14].

Our results can be applied to any linear recurrence sequences by using a similar method; for example, Lucas numbers, Pell numbers, Horadam numbers, generalized Fibonacci p-numbers.

The recurrence sequences determined by equalities (5.1), (5.2) and (5.3), (5.4) arise in a natural way when boundary value problems of type (4.1), (4.1) and (4.3), (4.4) are considered.

Let α = 1 2 ( x + x 2 + 4 ) and β = 1 2 ( x − x 2 + 4 ), then from the properties of second-order linear recurrence sequences, we have F n ( x ) = α n − β n x 2 + 4 and L n ( x ) = α n + β n.

Let α = 1 2 x + x 2 + 4 and β = 1 2 x - x 2 + 4, then from the properties of the second-order linear recurrence sequences we have F n x = α n - β n x 2 + 4 and  L n x = α n + β n.

It is clear that (F_{n}) is a second-order linear recurrence sequence.

The recurrence formula of { T n } is obtained by the properties of the third-order linear recurrence sequence.

From our theorem we know that if { b n } is a third-order linear recurrence sequence, then its Smarandache-Pascal derived sequence { T n } is also a third-order linear recurrence sequence.

Let (S_{n}) be an rth-order linear recurrence sequence satisfying the recursion (S_{n+r}=sum_{k=0}^{r-1}c_{k}S_{n+k} ).

Then for any positive integer d and T n + 1 = ∑ k = 0 n ( n k ) ⋅ P 2 d k + 1, we have the recurrence formula T n + 1 = ( 2 + P 2 d + 1 + P 2 d − 1 ) ⋅ ( T n − T n − 1 ), n ≥ 2. On the other hand, from our theorem, we know that if { b n } is a second-order linear recurrence sequence, then its Smarandache-Pascal derived sequence { T n } is also a second-order linear recurrence sequence.