Perfect square sequence 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)?
 A: 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 $b=10$).
A: 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 $a_n = \left( \frac{2(9m+1)^n+1}{3} \right)^2$.
A: $((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.
