In base 10, the sequence 49,4489,444889,... consists of all perfect squares.
Is this true for any other bases (greater than 10, of course)?
In base 10, the sequence 49,4489,444889,... consists of all perfect squares.
Is this true for any other bases (greater than 10, of course)?
No, it isn't.
The $n$'th term of your sequence in base $b$, if I understand correctly, is $1 + 8 \sum_{j=0}^{n-1} b^j + 4 \sum_{j=n}^{2n-1} b^j$. Consider the case $n=1$: $a_1 = 4 b + 9$. If that is a square, say $(2k+1)^2$ (since it is odd), we have $b = ((2k+1)^2 - 9)/4 = k^2 + k - 2$.
Now taking $n=3$: $a_3 = 4\,{k}^{10}+20\,{k}^{9}+4\,{k}^{8}-104\,{k}^{7}-64\,{k}^{6}+256\,{k}^{ 5}+120\,{k}^{4}-348\,{k}^{3}-24\,{k}^{2}+216\,k-71$. Consider $q = 2\,{k}^{5}+5\,{k}^{4}-{\frac {21}{4}}\,{k}^{3}-{\frac {103}{8}}\,{k}^{ 2}+{\frac {595}{64}}\,k+{\frac {891}{128}}$ (which comes from the Laurent series of $\sqrt{a_3/k^{10}}$ at $k=\infty$). It turns out that $a_3 - q^2 < 0$ for all real $k$, while $a_3 - (q - 1/128)^2 > 0$ for $k > 567.75$. Thus if $k$ is an integer $> 567$, $\sqrt{a_3}$ is between two numbers in of the form $\frac{\text{integer}}{128}$, and in particular is not an integer. Trying values of $k$ from 2 to 567, the only one that makes $a_3$ a square is $k=3$ (corresponding to $b=10$).
More generally, if $b = 9 m + 1$ and $r = 4 m$, the corresponding sequence $a_n = 1 + 2 r \sum_{j=0}^{n-1} b^j + r \sum_{j=n}^{2n-1} b^j$ consists of squares, namely $a_n = \left( \frac{2(9m+1)^n+1}{3} \right)^2$.
$((2*19^5+1)/3^2)=2724919437289,\ \ $ which converted to base $19$ is $88888GGGGH$. It doesn't work in base $13$ or $16$. In base $28$ it gives $CCCCCOOOOP$, where those are capital oh's (worth $24$).
This is because if we express $\frac{1}{9}$ in base $9a+1$, it is $0.aaaa\ldots$. So $\left (\frac{2(9a+1)^5+1}{3}\right)^2=\frac{4(9a+1)^10+4(9a+1)^5+1}{9}=$
$ (4a)(4a)(4a)(4a)(4a)(4a)(8a)(8a)(8a)(8a)(8a+1)_{9a+1}$
where the parentheses represent a single digit and changing the exponent from $5$ changes the length of the strings in the obvious way.