By 40 it's over 10^(100000)

A sequence of positive integers SS with S1=S2=1S_1=S_2=1 is further defined by Sk=Sk1×Sk2+1S_k=S_{k-1}\times S_{k-2}+1 for k>2.k > 2.

The second time in the sequence that the last three digits of SnS_{n} will equal the last three digits of Sn+1S_{n+1} will occur at Sn=A,S_{n=A}, whose last three digits are B.B. Find A+B.A+B.

Details and Assumptions\textbf{Details and Assumptions}
The first time is the trivial case at Sn=1.S_{n=1}.
For example, if the last three digits of both S193S_{193} and S194S_{194} were both 329,329, then A=193A=193 and B=329B=329 so the answer would be 522.522.


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