Abstract

Let (M_k) be the sequence of Mulatu numbers defined by M_0=4, M_1=1, M_k=M_(k-1)+M_(k-2) and (F_k) be the Fibonacci sequence given by the recurrence F_k=F_(k-1)+F_(k-2) with the initial conditions F_0=0, F_1=1 for k≥2. In this paper, we showed that all Mulatu numbers, that are concatenations of two Fibonacci numbers are 11,28. That is, we solved the equation M_k=〖10〗^d F_m+F_n, where d indicates the number of digits of F_n. We found the solutions of this equation as (k,m,n,d)∈{(4,2,2,1),(6,3,6,1)}. Moreover the solutions of this equation displayed as M_4=(F_2 F_2 ) ̅=11 and M_6=(F_3 F_6 ) ̅=28. Here the main tools are linear forms in logarithms and Baker Davenport basis reduction method.

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