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.
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.
It 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 =)