I come up with: $$\sum_{i=1}^{\left \lfloor{\frac{n+1}{2}}\right \rfloor}{{n-i+1}\choose{i}}$$
Which I think is true but is there a way to write it not as a sum?
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Try the Fibonacci numbers ... – Donald Splutterwit Jun 19 '17 at 19:35
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I mean to the number of components of the sum doesn't depend on $n$ – UfmdFkiF Jun 19 '17 at 19:35
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You can use inclusion-exclusion – Mickey Jun 19 '17 at 19:36
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@Mickey I see but what should I include/exclude? – UfmdFkiF Jun 19 '17 at 19:39
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Hint: if your number is $f(n)$, show that $f(n) = f(n-1) + f(n-2)$ by conditioning on the first member of the sequence. – Robert Israel Jun 19 '17 at 19:44
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$r_1=n\cdot(1),m\cdot(0)$||||| $r_2=n+m\cdot(1),n\cdot(0)$||||| $r_3=2n+m\cdot(1),n+m\cdot(0)$ – Jacob Claassen Jun 19 '17 at 19:57
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@RobertIsrael Ok but How to even come up with the idea of using the Fibonacci sequence? – UfmdFkiF Jun 19 '17 at 20:20
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Conditioning on the first term is often a good strategy. Once you get the recursion, you recognize it as the Fibonacci recursion and then you're done. – Robert Israel Jun 19 '17 at 21:04
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For $n=1,2,3,...$ the answer is $2$, $3$, $5$, $8$, etc. Maybe a well-known sequence?
Angina Seng
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2OEIS says that there are 825 well-known sequences with 2, 3, 5, 8 $\ddot\smile$. – Smylic Jun 19 '17 at 19:59
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