Does $ \{na\}$ mean $ na$ ? If so, it depends on the value of $ a$ whether a sequence with the expected property exists or not. For trivial cases (i) When $ a\leq 0$ we have such a sequence $ 1,2,4,7,\dots$ (ii) There exists no such sequence if $ a\geq \frac{1}{10}$ We are left with finding the interval $ \in (0,\frac{1}{10})$ so we can find at least one sequence when $ a$ lies interval. Observing that $ 1,2,4,7,\dots$ is the slowest growing sequence with the expected property, \[ 7\alpha<\frac{1}{10}\] Hence $ a$ has to lie in the interval $ (0,\frac{1}{70})$

srikanth wrote: Does $ \{na\}$ mean $ na$ ? No it means the fractional part of $ na$, i.e. $ \{na\} = na - \lfloor na \rfloor$.

Hmm. It looks like in the case where $ a<1/10$ we can get the upper bound of 3 (havn't checked for rational though) however $ a>1/10$ seems not easy. I tried pigeonhole principle for fractional parts of $ a,2a,\ldots,na$ and similar techniques but failed. Can anyone help?

A friend of mine suggested that $ \frac1{10}$ could be replaced by virtually any constant $ c$ (we can assume $ c<1$) and the result still holds. We tested this on several different pairs $ (a,c)$ and his hypothesis was confirmed. Moreover, a result which is probably easy to show is that if we have three different differences, then the sum of the two smaller ones equals the third.

Let us first note that the number $\alpha$ can be considered to lie in the interval $(0, 1)$. Indeed, when any integer is added to $\alpha$, the values of $\{n\alpha\}$ do not change. It will be useful for us to imagine the numbers $\{n\alpha\}$ as points on a circle of unit length, on which arcs of length $\alpha$ are successively laid off from point 0. In such an interpretation we will be concerned with the mutual arrangement of points on a given arc of length $1/10$. Let us call the transition from the point $\{n_k\alpha\}$ to the point $\{n_{k+1}\alpha\}$ a shift to the right if $\{n_{k+1}\alpha\} > \{n_{k}\alpha\}$, and to the left if $\{n_{k+1}\alpha\} < \{n_k\alpha\}$ (case $\{n_{k+1}\alpha\} = \{n_k\alpha\}$ is trivial). We will call the magnitude of the shift (or just "shift of ...") the difference $|\{n_{k+1}\alpha\} - \{n_k\alpha\}|$ and its number of steps the difference $n_{k+1} - n_k$. Note first that two consecutive shifts in the same direction are always equal. Indeed, if, for example, $\{n_k\alpha\} < \{n_{k+1}\alpha\} < \{n_{k+2}\alpha\}$, then $n_k < m = n_k + n_{k+2} - n_{k+1} < n_{k+2}$ and $\{m\alpha\} = \{n_k\alpha\} + \{n_{k+2}\alpha\} - \{n_{k+1}\alpha\} < 1/10$, i.e., $m = n_{k+1}$ and $n_{k+2} - n_{k+1} = n_{k+1} - n_k$. Choose among all possible shifts the right shift with the smallest number of steps $m_1$ of $\Delta_1$ and the left shift with the smallest number of steps $m_2$ of $\Delta_2$. We may assume that $m_1 < m2$. We prove the two lemmas: lemma 1) if $\Delta_1 +\Delta_2 \geq 1/10$ (i think it is true always) and for some $n$ the fractional part $\{n\alpha\}$ lies in the interval $[1/10 - \Delta_1, \Delta_2)$, then there exists a shift of $|\Delta_2 - \Delta_1|$ with number of steps $m_1 + m2$; lemma 2) there are no shifts with a number of steps no more than $m_1 + m_2$, except for the three (or possibly two) mentioned above. Proof of lemma 2. Let such a shift of $\Delta$ with number of steps $m$ exist. If it is a shift to the right, then $\Delta < \Delta_1$. Let us find $n_k$ such that $\{n_k\alpha\} < 1/10 - \Delta_1$ (i.e., one to which the shift by $\Delta_1$ applies). Then $\{(n_k + m)\alpha\} = \{n_k\alpha\} + \Delta < \{n_k\alpha\} + \Delta_1 = \{(n_k + m_1)\alpha\}$. Since $m$ cannot be less than $m_1$, the transition from $\{(n_k +m_1)\alpha\}$ to $\{(n_k +m)\alpha\}$ contains a shift to the left and less than $m_2$ steps, which is impossible. If the shift by $\Delta$ is a shift to the left, then $\Delta < \Delta_2$ and $m > m_2$. Taking $n_k$ such that $\{n_k\alpha\} > \Delta_2$, we obtain that the transition from $\{(n_k + m2)\alpha\}$ to $\{(n_k + m)\alpha\}$ contains a shift to the right and less than $m_1$ steps -- again a contradiction. Proof of lemma 1. Take a natural $n$ for which $\{n\alpha\}$ lies on the half-interval $[1/10 -\Delta_1, \Delta_2)$. Obviously, $\{(n+m_1 +m_2)\alpha\} = \{n\alpha\} + \Delta_1 - \Delta_2\in (0, 1/10)$. On the other hand, at no natural $k < m_1 + m_2$ does the fractional part $\{(n + k)\alpha\}$ lie in the interval $(0, 1/10)$ -- at $k = m_1$ and $k = m_2$ by the assumption for the number $\{n\alpha\}$, and at the other $k$ by lemma 2, which has already been proved. The statement of the problem immediately follows from lemmas 1 and 2. Indeed, if $\Delta_1 +\Delta_2 \geq 1/10$ and for some $n$ the fractional part $\{n\alpha\}$ lies on the interval $[1/10- \Delta_1, \Delta_2)$, then from points with fractional parts smaller than $1/10-\Delta_1$ we have shift by $\Delta_1$, from points with fractional parts greater than $\Delta_2$ are shifted by $-\Delta_2$ and the other ones are shifted by |\Delta_2 - \Delta_1| (both shifts with less number of steps are impossible). The case where no such $n$ exists differs only in the fact that there are no points of the last type. Note that we have proved even somewhat more than required in the problem condition: it turns out that if the differences $n_{k+1} - n_k$ take three different values, then one of these values is equal to the sum of the other two.

micliva wrote: Let us first note that the number $\alpha$ can be considered to lie in the interval $(0, 1)$. Indeed, when any integer is added to $\alpha$, the values of $\{n\alpha\}$ do not change. It will be useful for us to imagine the numbers $\{n\alpha\}$ as points on a circle of unit length, on which arcs of length $\alpha$ are successively laid off from point 0. In such an interpretation we will be concerned with the mutual arrangement of points on a given arc of length $1/10$. Let us call the transition from the point $\{n_k\alpha\}$ to the point $\{n_{k+1}\alpha\}$ a shift to the right if $\{n_{k+1}\alpha\} > \{n_{k}\alpha\}$, and to the left if $\{n_{k+1}\alpha\} < \{n_k\alpha\}$ (case $\{n_{k+1}\alpha\} = \{n_k\alpha\}$ is trivial). We will call the magnitude of the shift the difference $\{n_{k+1}\alpha\} - \{n_k\alpha\}|$ and its number of steps the difference $n_{k+1} - n_k$. Note first that two consecutive shifts in the same direction are always equal. Indeed, if, for example, $\{n_k\alpha\} < \{n_{k+1}\alpha\} < \{n_{k+2}\alpha\}$, then $n_k < m = n_k + n_{k+2} - n_{k+1} < n_{k+2}$ and $\{m\alpha\} = \{n_k\alpha\} + \{n_{k+2}\alpha\} - \{n_{k+1}\alpha\} < 1/10$, i.e., $m = n_{k+1}$ and $n_{k+2} - n_{k+1} = n_{k+1} - n_k$. Choose among all possible shifts the right shift with the smallest number of steps $m_1$ of $\Delta_1$ and the left shift with the smallest number of steps $m_2$ of $\Delta_2$. We may assume that $m_1 < m2$. We prove the two lemmas: lemma 1) if $\Delta_1 +\Delta_2 \geq 1/10$ (i think it is true always) and for some $n$ the fractional part $\{n\alpha\}$ lies in the interval $[1/10 - \Delta_1, \Delta_2)$, then there exists a shift of $|\Delta_2 - \Delta_1|$ with number of steps $m_1 + m2$; lemma 2) there are no shifts with a number of steps no more than $m_1 + m_2$, except for the three (or possibly two) mentioned above. Proof of lemma 2. Let such a shift of $\Delta$ with number of steps $m$ exist. If it is a shift to the right, then $\Delta < \Delta_1$. Let us find $n_k$ such that $\{n_k\alpha\} < 1/10 - \Delta_1$ (i.e., one to which the shift by $\Delta_1$ applies). Then $\{(n_k + m)\alpha\} = \{n_k\alpha\} + \Delta < \{n_k\alpha\} + \Delta_1 = \{(n_k + m_1)\alpha\}$. Since $m$ cannot be less than $m_1$, the transition from $\{(n_k +m_1)\alpha\}$ to $\{(n_k +m)\alpha\}$ contains a shift to the left and less than $m_2$ steps, which is impossible. If the shift by $\Delta$ is a shift to the left, then $\Delta < \Delta_2$ and $m > m_2$. Taking $n_k$ such that $\{n_k\alpha\} > \Delta_2$, we obtain that the transition from $\{(n_k + m2)\alpha\}$ to $\{(n_k + m)\alpha\}$ contains a shift to the right and less than $m_1$ steps -- again a contradiction. Proof of lemma 1. Take a natural $n$ for which $\{n\alpha\}$ lies on the half-interval $[1/10 -\Delta_1, \Delta_2)$. Obviously, $\{(n+m_1 +m_2)\alpha\} = \{n\alpha\} + \Delta_1 - \Delta_2\in (0, 1/10)$. On the other hand, at no natural $k < m_1 + m_2$ does the fractional part $\{(n + k)\alpha\}$ lie in the interval $(0, 1/10)$ -- at $k = m_1$ and $k = m_2$ by the assumption for the number $\{n\alpha\}$, and at the other $k$ by lemma 2, which has already been proved. The statement of the problem immediately follows from lemmas 1 and 2. Indeed, if $\Delta_1 +\Delta_2 \geq 1/10$ and for some $n$ the fractional part $\{n\alpha\}$ lies on the interval $[1/10- \Delta_1, \Delta_2)$, then from points with fractional parts smaller than $1/10-\Delta_1$ we have shift by $\Delta_1$, from points with fractional parts greater than $\Delta_2$ are shifted by $-\Delta_2$ and the other ones are shifted by |\Delta_2 - \Delta_1| (both shifts with less number of steps are impossible). The case where no such $n$ exists differs only in the fact that there are no points of the last type. Note that we have proved even somewhat more than required in the problem condition: it turns out that if the differences $n_{k+1} - n_k$ take three different values, then one of these values is equal to the sum of the other two. sorry for asking but what does "delta notation"stand for?magnitude of shift?or just represents a certain shift?

blackmetalmusic wrote: sorry for asking but what does "delta notation"stand for?magnitude of shift?or just represents a certain shift? Yep, magnitude.