A solution of the following difference equation is investigated:xn+1=xn−k+11+xnxn−1…xn−k,n=0,1,2,…$$ {x}_{n+1}=\frac{x_{n-\left(k+1\right)}}{1+{x}_n{x}_{n-1}\dots {x}_{n-k}},n=0,1,2,\dots $$where x−(k+1); x−k; : : : ; x−1; x0 ???? (0;∞) and k = 0; 1; 2; : :… Click to show full abstract
A solution of the following difference equation is investigated:xn+1=xn−k+11+xnxn−1…xn−k,n=0,1,2,…$$ {x}_{n+1}=\frac{x_{n-\left(k+1\right)}}{1+{x}_n{x}_{n-1}\dots {x}_{n-k}},n=0,1,2,\dots $$where x−(k+1); x−k; : : : ; x−1; x0 ???? (0;∞) and k = 0; 1; 2; : : : .
               
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