What comes next in this sequence?
4, 3, 9, 5, 19, 9, 39, 17, 79, 33, ?
It didn't immediately jump out at me, but ended up not too challenging and thought some may enjoy it.
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Sign up to join this communityIt is:
4, 3, 9, 5, 19, 9, 39, 17, 79, 33, 159, 65, 319, 129
Because:
The differences between every other number are:
5, 2, 10, 4, 20, 8, 40, 16
So the odd entries begin with $4$ and add $5\times2^n$ to each.
The even entries begin with $3$ and add $2^n$ to each.
In appropriate math notation, thanks to f'':
$2x\,\text{th}$ term is $2^x+1$ and the $2x+1\,\text{th}$ term is $5⋅2^x−1$.
The next numbers in sequence are
159, 65, 319, ...
The formula of the sequence is:
$(2^{\frac{n}{2}}) \cdot (2^{\frac{3}{2}}+2^{-\frac{1}{2}})^{nmod2}+(-1)^{n}$, where $n$ is the nth term and $nmod2$ is n modulo 2
The formula is deduced using the following logic:
To get the odd terms (where n = 1, 3, 5, ...), the following formula is used:
$2^{(\frac{n+1}{2}+1)} + 2^{(\frac{n+1}{2}-1)} - 1$
$=(2^{\frac{n}{2}})(2^{\frac{3}{2}}+2^{-\frac{1}{2}}) - 1$
To get the even terms (where n = 2, 4, 6, ...). the following formula is used:
$2^{\frac{n}{2}}+1$
There are only two differences in the above two formulas, one is $(2^{\frac{3}{2}}+2^{-\frac{1}{2}})$ and the other is the last constant $1$. To combine those two formulas, these two differences need to be addressed
1. As can be deduced from the above two formulas, $(2^{\frac{3}{2}}+2^{-\frac{1}{2}})$ is needed to obtain the odd terms while it is not needed to obtain the even terms. Hence a power of $nmod2$ is added so that $(2^{\frac{3}{2}}+2^{-\frac{1}{2}})^{nmod2} = 1$ when $n$ is even
2. As for the last constant, using $(-1)^n$ should suffice
After combination, the final formula is obtained: $(2^{\frac{n}{2}}) \cdot (2^{\frac{3}{2}}+2^{-\frac{1}{2}})^{nmod2}+(-1)^{n}$
I hope that my logic of deduction is easy enough to understand. Do comment on unclear parts =)