Well I got it from a competitive exam paper from 2012. $$ { } $$ Let $a_n$ be the number of all $n$ - digit positive integers formed by digits $0,1$ or both such that no consecutive digits in them are $0$. Let $b_n =$ number of such integers ending with $1$ and $c_n =$ number of such integers ending with $0$. $$ { } $$ Then I made a table: $$1$$ $$10,11$$ $$101, 110, 111$$ $$1010, 1011, 1101, 1110, 1111$$ $$10101, 10110, 10111, 11010, 11011, 11110, 11111$$ $$....\text {etc.} $$ While $c_n=F_n$ is clearly the Fibonacci sequence, $b_n=F_{n+1}$ and $a_n=F_{n+2}$. How? Is there a way to prove it? I thought perhaps we could use combinatorics but, just a thought, no way out. Any help/suggestions are welcome. Thanks!
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There are many duplicates. Searching for binary consecutive will find many of them. – Ross Millikan Nov 25 '19 at 05:43
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@Ross Millikan , thanks for help. – Awe Kumar Jha Nov 25 '19 at 05:48