again as i was looking at older pages this one caught my attention and in my opinion this should have been solved long ago. i claim the answer is the set of those $n$ for which $\dfrac{n(n+1)}{2}$ is a perfect square(of course there are infintely many such $n$ that can be found by the so-called Pell's method or whatever) to start with let $K_n=\{k\le n|\dfrac{k(k+1)}{2} = \square\}$.(i.e is a square) suppose $k,k+1\in K_n$. Then $k^2+k=2r^2,k^2+3k+2=2s^2$ where $r,s$ are positive integers. Then $k+1=s^2-r^2$. Plugging this back into the two equations and dividing, we have $\dfrac{s^2-r^2-1}{s^2-r^2+1}=\dfrac{r^2}{s^2}$ so that we have $(s^2-r^2)^2=(s^2+r^2)$. Now we know that $(s^2-r^2)^2+(2rs)^2=(s^2+r^2)^2$ so that we have $(r^2+s^2)(r^2+s^2-1)=(2rs)^2$ and that is a contradiction since product of two consecutive integers cannot be a perfect square. hence $K_n$ has 'gaps' and we an use these gaps to get a sequence without any partial squares appearing. start with the word $21$ and add all consecutive integers till you hit an element $k$ of $K_n$; then we throw in $k+1$ before $k$ and then $k$. for example: $2134567\underline{98}\ldots$. we just exchange at the occurence of the squares and by what we have just seen, this exchange will not create any partial squares till the integer $k+1$ and we can carry on the same way.

seshadri wrote: again as i was looking at older pages this one caught my attention and in my opinion this should have been solved long ago. i claim the answer is the set of those $n$ for which $\dfrac{n(n+1)}{2}$ is a perfect square(of course there are infintely many such $n$ that can be found by the so-called Pell's method or whatever) to start with let $K_n=\{k\le n|\dfrac{k(k+1)}{2} = \square\}$.(i.e is a square) suppose $k,k+1\in K_n$. Then $k^2+k=2r^2,k^2+3k+2=2s^2$ where $r,s$ are positive integers. Then $k+1=s^2-r^2$. Plugging this back into the two equations and dividing, we have $\dfrac{s^2-r^2-1}{s^2-r^2+1}=\dfrac{r^2}{s^2}$ so that we have $(s^2-r^2)^2=(s^2+r^2)$. Now we know that $(s^2-r^2)^2+(2rs)^2=(s^2+r^2)^2$ so that we have $(r^2+s^2)(r^2+s^2-1)=(2rs)^2$ and that is a contradiction since product of two consecutive integers cannot be a perfect square. hence $K_n$ has 'gaps' and we an use these gaps to get a sequence without any partial squares appearing. start with the word $21$ and add all consecutive integers till you hit an element $k$ of $K_n$; then we throw in $k+1$ before $k$ and then $k$. for example: $2134567\underline{98}\ldots$. we just exchange at the occurence of the squares and by what we have just seen, this exchange will not create any partial squares till the integer $k+1$ and we can carry on the same way. It is not clear how we can create sequance $2134567\underline{98} \ldots$.Words "we can carry in the same way" doesn't make sense.

The idea is that he will create a sequence $a_i$ such that if $\frac{n(n+1)}{2}$ is a square then the only square in the sums $a_1,...,a_1+...+a_n$ is $a_1+a_2+...+a_n=\frac{n(n+1)}{2}$, and $a_n=n$, and if that number is not a square then none of the sums will have a square. This can be done inductively. Base case is clear. Suppose for $k$ such a sequence exist, for $k+1$, if $\frac{k(k+1)}{2}$ is not a square then just add $a_{k+1}=k+1$. Then none of sums $a_1,a_1+a_2,...,a_1+...+a_k$ is a perfect square, and $(a_1,a_2....a_{k+1})$ is a permutation of $1$ to $k+1$. (and if $\frac{(k+1)(k+2)}{2}$ is a square it has $a_{k+1}=k+1$). If $\frac{k(k+1)}{2}$ is a square then change $a_k=k$ to $a_k=k+1$ and then let $a_{k+1}=k$. Clearly now there cannot be any square in the sums ($a_1,...,a_1+...a_k$(which is some square minus $1$) is clear, and $a_1+...+a_{k+1}$ is not a square from elements of $K_n$ are not consecutive.). By induction we can construct the necessary sequences.