We solve the problem in the general case: A student is preparing during $n$ days for a contest. The student alternates days of training with days of rest, such that there are not $3$ consecutive days of the same kind. Determine the number of possible configurations $C(n)$ of the preparing days for the contest. Denote: $T=training;R=rest$; $S(n)=$ the number of configurations of the $n$ preparing days such that the last $2$ days are of the same kind ($TT$ or $RR$); $D(n)=$ the number of configurations of the $n$ preparing days such that the last $2$ days are of different kind ($TR$ or $RT$). Results: $C(n)=S(n)+D(n);S(n+1)=D(n);D(n+1)=C(n)$. Let be the Fibonacci sequence $F_0=F_1=1; F_{k+1}=F_k+F_{k-1}, \forall k\in\mathbb{N}$. For $n=2$: $S(2)=2=2F_0\; (TT;RR);\;D(2)=2=2F_1\; (TR;RT)\;;S(2)=2(F_0+F_1)=2F_2$. Results for $n=3$: $S(3)=D(2)=2F_1; D(3)=C(2)=2F_2; C(3)=2(F_1+F_2)=2F_3$. By induction: $S(n)=2F_{n-2};D(n)=2F_{n-1};C(n)=2F_n$. Hence: $C(14)=2F_{14}=2\cdot610=1220$.