An element x in a noncommutative ring R is strongly nilpotent if any chain $x_1=x, x_2, ... $, with $x_{n+1}\in x_n R x_n$ terminates at zero. It becomes clear why this is a good definition once one has shown that the set of all such elements is the semiprime radical (the intersection of all prime ideals). However, one can ask if this is equivalent to another definition (which at the first glance seems more natural): x is naively strongly nilpotent if for all $a_1,a_2,..$ in $R$ we have $x a_1 x a_2 .. x a_n x =0$ for some $n$. "Naively strongly nilpotent" implies strongly nilpotent; is the converse true in general? [Probably not, but what's a counterexample?]
Consider the infinite word $W=xx_1xx_2xx_3x....$. Let $R=R(W)$ be the ring which consists of all integral linear combinations of finite subwords of $W$. The product of two subwords $u,v$ is either the concatenation $uv$ if $uv$ is a subword of $W$ or $0$ otherwise. It is clearly an associative ring. The element $x$ in $R(W)$ is not naively strongly nilpotent but is strongly nilpotent. One can also get a finitely generated ring this way. Just consider the word $W=xyxy^2x...$. The ring $R(W)$ is 2generated, $x$ is strongly nilpotent but not "naive". Here is the proof. The fact that $x$ is not "naively strongly nilpotent" follows from the fact that the words $x,xyx, xyxy^2x,...$ are subwords of $W$ and hence not equal to 0 in $R(W)$. Consider any sequence $t_0=x, t_1\in xR(W)x, t_2\in t_1R(W)t_1,...$, $t_i\ne 0$. Let $t_i=\sum_{j=1}^{n_i} w^{(i)}_j$, where each $w^{(i)}_j$ is a subword of $W$. Then each $w^{(i)}_j$ starts and ends with $x$ and is of length at least 2, each $t_p$, $p > 1$, is a linear combination of words which start with some $w_k^{(1)}$ and end with some $w^{(1)}_l$. But this implies that the lengths of the summands of $t_p$ are bounded (since a long subword of $W$ cannot end with a short word of the form $x...x$), a contradiction. Thus $x$ is strongly nilpotent. 


However Roman may still have a point: the strong nilpotents depend on the naively strong nilpotents in the sense that if there are no naively strong nilpotents, then there are no strong nilpotents. Proof: If there are no (nonzero) naively strong nilpotents, then there are no nilpotent ideals. Hence for any $x\ne0$, one can always find $0\ne x_{n+1}\in x_nRx_n\subseteq\langle x_n\rangle^2$, $x_0=x$. So $x$ is not strongly nilpotent. Strong nilpotents $a$ are such that 'fractal'like words $ax_1ax_2ax_1ax_3ax_1ax_2ax_1a$ are eventually 0. This seems either a deep or an unnatural phenomenon that can be simplified. By unnatural is meant that if one defines a strong3nilpotent similarly to a strong nilpotent but with $x_{n+1}\in x_nRx_nRx_n\subseteq\langle x_n\rangle^3$, then the two definitions actually agree. 

