(2) Yes. Let us take $a_0 = 1$, $a_1 = 1/\sqrt{2}$, and impose $a_{n+1}^2 = 2a_na_{n+2}$ for $n\geq 0$. Denote $x_k = \ln a_k$ for all $k$, so $x_{n+2} - 2x_{n+1} + x_n + \ln 2 = 0$. Wriye also $x_{n+3} - 2x_{n+2} + x_{n+1} + \ln 2 = 0$, and subtract the two, to get $x_{n+3} - 3x_{n+2} + 3x_{n+1} - x_n = 0$. The characteristic polynomial for this recurrence relation is $(\lambda-1)^3 = \lambda^3 - 3\lambda^2 + 3\lambda - 1 = 0$, therefore the general solution is $x_n = \alpha + \beta n + \gamma n^2$. Determining the coefficients $\alpha, \beta, \gamma$ from the values $x_0 = 0$, $x_1 = -\dfrac {1} {2}\ln 2$ and $x_2 = 2x_1 - x_0 - \ln 2 = -2\ln 2$ yields $\alpha = \beta = 0$ and $\gamma = -\dfrac {1} {2}\ln 2$, thus $x_n = -\dfrac {n^2} {2}\ln 2$ and $a_n = 2^{-n^2/2}$. (1) Most likely the answer No can be obtained from similar considerations, comparing the "inequality" relation with the "equality" relation imposed above (and which has known solutions), and reaching the conclusion that the sequence should be ultimately decreasing, which is impossible for a sequence of positive integers. I lack the drive to do it in full detail, since I'm pretty sure it works.

mavropnevma wrote: (2) Yes. Let us take $a_0 = 1$, $a_1 = 1/\sqrt{2}$, and impose $a_{n+1}^2 = 2a_na_{n+2}$ for $n\geq 0$. Denote $x_k = \ln a_k$ for all $k$, so $x_{n+2} - 2x_{n+1} + x_n + \ln 2 = 0$. Wriye also $x_{n+3} - 2x_{n+2} + x_{n+1} + \ln 2 = 0$, and subtract the two, to get $x_{n+3} - 3x_{n+2} + 3x_{n+1} - x_n = 0$. The characteristic polynomial for this recurrence relation is $(\lambda-1)^3 = \lambda^3 - 3\lambda^2 + 3\lambda - 1 = 0$, therefore the general solution is $x_n = \alpha + \beta n + \gamma n^2$. Determining the coefficients $\alpha, \beta, \gamma$ from the values $x_0 = 0$, $x_1 = -\dfrac {1} {2}\ln 2$ and $x_2 = 2x_1 - x_0 - \ln 2 = -2\ln 2$ yields $\alpha = \beta = 0$ and $\gamma = -\dfrac {1} {2}\ln 2$, thus $x_n = -\dfrac {n^2} {2}\ln 2$ and $a_n = 2^{-n^2/2}$. There is tiny flaw here. If $2|n$, $a_n$ is not irrational. We actually could let $a_n=\sqrt{3} 2^{-n^2/2}$ mavropnevma wrote: (1) Most likely the answer No can be obtained from similar considerations, comparing the "inequality" relation with the "equality" relation imposed above (and which has known solutions), and reaching the conclusion that the sequence should be ultimately decreasing, which is impossible for a sequence of positive integers. I lack the drive to do it in full detail, since I'm pretty sure it works. This is right. $\frac{a_{n+2}}{a_{n+1}}\le \frac 12 \frac{a_{n+1}}{a_{n}}\le 2^{-n-1}\frac{a_1}{a_0}$. So eventually $a_n$ will be strictly decreasing, which is impossible.

xxp2000 wrote: There is tiny flaw here. If $2\mid n$, $a_n$ is not irrational. We actually could let $a_n=\sqrt{3} \cdot 2^{-n^2/2}$ Oh, I did not lose any sleep on that I took the true meaning of (2) to be geared not towards pure irrationality of the terms, but to the fact that dropping the integer requirement we can actually find a solution. As pointed by you, there is a quick fix, if we insist on pure irrationality.

mavropnevma wrote: xxp2000 wrote: There is tiny flaw here. If $2\mid n$, $a_n$ is not irrational. We actually could let $a_n=\sqrt{3} \cdot 2^{-n^2/2}$ Oh, I did not lose any sleep on that I took the true meaning of (2) to be geared not towards pure irrationality of the terms, but to the fact that dropping the integer requirement we can actually find a solution. As pointed by you, there is a quick fix, if we insist on pure irrationality. No big deal! But if we assume 2) means "not pure irrationality", we should also assume 1) means "not pure integer", then 1) actually has an answer.

is we define $\{b_n\}$ like this : $b_n=a_{n+1}/a_{n}$ then we get $b_n\ge2b_{n+1}$ so put $b_i=1/2^i$ and put $a_1=r$ which r is irrational .and it works i guess...