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In mathematics (including combinatorics, linear algebra, and dynamical systems), a linear recurrence with constant coefficients [1]: ch. 17 [2]: ch. 10 (also known as a linear recurrence relation or linear difference equation) sets equal to 0 a polynomial that is linear in the various iterates of a variable—that is, in the values of the elements of a sequence.
If the {} and {} are constant and independent of the step index n, then the TTRR is a Linear recurrence with constant coefficients of order 2. Arguably the simplest, and most prominent, example for this case is the Fibonacci sequence , which has constant coefficients a n = b n = 1 {\displaystyle a_{n}=b_{n}=1} .
A famous example is the recurrence for the Fibonacci numbers, = + where the order is two and the linear function merely adds the two previous terms. This example is a linear recurrence with constant coefficients, because the coefficients of the linear function (1 and 1) are constants that do not depend on .
F(n) = F(n − 1) + F(n − 2) together with the initial values F(0) = 0 and F(1) = 1. The Skolem problem is named after Thoralf Skolem, because of his 1933 paper proving the Skolem–Mahler–Lech theorem on the zeros of a sequence satisfying a linear recurrence with constant coefficients. [2]
The sequence is determined by the linear recurrence equation with polynomial coefficients = () + and the initial values () =, =. Applying an algorithm to find hypergeometric solutions one can find the general hypergeometric solution y ( n ) = c 2 n n ! {\displaystyle y(n)=c\,2^{n}n!} for some constant c {\textstyle c} .
which implies the first-order linear recurrence with constant coefficients y n + 1 = y n ( 1 − h κ ) . {\\displaystyle y_{n+1}=y_{n}(1-h\\kappa ).} Given y ( 0 ) = y 0 {\\displaystyle y(0)=y_{0}} , the sequence satisfying that recurrence is the geometric progression
is constant-recursive because it satisfies the linear recurrence = +: each number in the sequence is the sum of the previous two. [2] Other examples include the power of two sequence 1 , 2 , 4 , 8 , 16 , … {\displaystyle 1,2,4,8,16,\ldots } , where each number is the sum of twice the previous number, and the square number sequence 0 , 1 , 4 ...
If one does not end if a solution is found it is possible to combine all hypergeometric solutions to get a general hypergeometric solution of the recurrence equation, i.e. a generating set for the kernel of the recurrence equation in the linear span of hypergeometric sequences. [1] Petkovšek also showed how the inhomogeneous problem can be solved.