The answer is all $M>1$. Clearly, $M=1$ fails, as the sequence is just going to be all $\frac{3}{2}$. To show that all $M>1$ work, we induct on $v_2(M-1)$. If $v_2(M-1)=0$, then $M$ is even, and we have that $a_1=M(M+\frac{1}{2})$ is an integer as desired. If $v_2(M-1)=k>0$, then $a_1=M(M+\frac{1}{2})=\frac{2M^2+M-1}{2}+\frac{1}{2}=N+\frac{1}{2}$, and $v_2(N-1)=v_2(\frac{2M^2+M-3}{2})=v_2(\frac{(2M+3)(M-1)}{2})=k-1$, so we can shift the sequence and reduce to the case of $v_2(M-1)=k-1$. Hence, $a_{k+1}$ will be an integer as desired.

The answer is all positive integers $M\ge 2$. Call an positive integer $M$ good if such term exist, and bad otherwise. We prove that $1$ is the only bad positive integer. Notice that for any $k$, $\frac{2k+1}{2}\lfloor {\frac{2k+1}{2}} \rfloor=\frac{k(2k+1)}{2}\in \mathbb{N}\iff k\enspace\text{is even.}$ So all even numbers are good. Now consider $k$ be an odd integer, then $$\frac{k(2k+1)}{2}=\frac{2(k^2+\lfloor \frac{k}{2}\rfloor)+1}{2}$$So $k$ is good $\iff$ $k^2+\lfloor \frac{k}{2}\rfloor$ is good. Now notice that if $k$ is in the form of $4p+3$, then $(4p+3)^2+\lfloor \frac{4p+3}{2}\rfloor$ is even, so $4p+3$ is good for all $p$. Now suppose $k$ is in the form $2^{i+1}p+(2^i+1)$, and suppose all numbers in the form $2^ip+(2^{i-1}+1)$ is good. Then notice that $$(2^{i+1}p+(2^i+1))^2+\lfloor\frac{2^{i+1}p+(2^i+1)}{2} \rfloor\equiv 1+2^{i-1}\pmod {2^i}$$which is in the form of $2^ip+(2^{i-1}+1)$, which is good. So all numbers in the form of $2^{i+1}p+(2^i+1)$ is good as well. Note that any odd integer $k$ at least $2$ can be written as $k=2^{i+1}p+(2^i+1)$ for some suitable $i, p$, because we can take $i=v_2(k-1)$ and then $p$ follows directly. So this proves that all $M\ge 2$ are good. For $M=1$, then $a_i=\frac{3}{2}$ for all $i$, so $M=1$ is bad. So the answer is all positive integers $M\ge 2$.

This was also German TSTST #4

Just notice that if we assume to the contrary that this is not true then the greatest integers less than each of the terms should be odd which is easily proven to be true if and only if the sequence is constant.

I thinck this works : Call a number bad if it does not respect the statement . Then prove by induction that all candidates for bad numbers must be of the form $2^n + 1$ . So now if we have a bad number $M$ then $M - 1$ is divizible with every power of $2$ so $M = 1$

I made a typo : there should be $2^n k + 1$

The answer is all $M>1$. We induct on $v_2(M-1)$. If $v_2(M-1)=0$, i.e. $M$ is even, then $\lfloor a_0 \rfloor \equiv 0 \pmod{2}$, so we are clear on $a_1$. Suppose the sequence has at least one integer term if $v_2(M-1)=k-1$. We prove the statement when $v_2(M-1)=k$. Let $M=2^kn+1$. Then $a_1=(2^kn+1)^2 + \frac{2^kn+1}{2} = (2^{2k}n^2+2^{k+1}n+2^{k-1}n)+\frac{1}{2}$. Now note that $v_2(2^{2k}n^2+2^{k+1}n+2^{k-1}n)=k-1$, so by shifting the indices we are good to go. So pretty much the first integer comes up is $a_{v_2(M-1)+1}$ I guess?

For $M=1$ we obtain $a_k=\frac{3}{2}\not\in\mathbb{Z},\forall k\ge 0$,so $1$ does not satisfy the given condition. We will now prove that any integer $M>1$ satisfies the given condition. Suppose that there exists an integer $M>1$ such that the sequence $(a_k)$ doesn't contain any integer term.In this case it's obvious that all the terms of the sequence will be of the form $t+\frac{1}{2},t\in\mathbb{Z}$. $a_1=M^2+\frac{M}{2}\not\in\mathbb{Z}$,so $M\equiv 1\pmod{2}$ i.e. $\lfloor a_0\rfloor\equiv 1\pmod{2}$.By looking at $a_1$ and $a_2$,we obtain,in a similar fashion,that $\lfloor a_1\rfloor\equiv 1\pmod{2}$ i.e. $M^2+\frac{M-1}{2}\equiv 1\pmod{2}$ or $M\equiv 1\pmod{4}$ i.e. $\lfloor a_0\rfloor\equiv 1\pmod{4}$. By the same argument as above we conclude that $\lfloor a_1\rfloor\equiv 1\pmod{4}$ or $M^2+\frac{M-1}{2}\equiv 1\pmod{4}$,which implies that $M\equiv 1\pmod{8}$(we used $M\equiv 1\pmod{4}$). By repeating this over and over again we obtain that $M\equiv 1\pmod{2^n},\forall n\in\mathbb{N}$,so $M=1$,contradiction! Hence the answer:All positive integers $M>1$ satisfy the given condition.

Just a gist: 1)Clearly, there is no integer term in the sequence for $M=1$. 2) For $M >1$, let $2^{\alpha}||(M-1) $. Then $a_{\alpha +1} $ is an integer. 3) To prove the second point, note that if $a_{t} = 2^{j}*k_{t} + 1.5 $ for some positive integer $j $ and odd natural $k_{t}$, then $a_{t+1} $ is of the form $2^{j-1}*k_{t+1} +1.5$ for odd natural $k_{t+1} $.

Had I solved this one in December, then I would've made it into the German IMO TST group but well I failed back then. Now I've come back and have found a solution in quite a short while. Would've been more use to me, if that was 9 months ago, though. We'll prove that all $M \geq 2$ satisfy the conditions. It is easy to show that $M=1$ is not a solution. So we'll now prove that all $M \geq 2$ actually are solutions. Assume the contrary. So suppose there is a $M \geq 2$, such that all numbers $a_0,a_1,a_2,\dots$ are not integers. Since $a_0=\tfrac{2M+1}{2}$ and $a_{k+1}=a_k \cdot \lfloor a_k \rfloor$ only multiplies the previous number of the sequence with a integer, all numbers $a_0,a_1,a_2,\dots$ are of the form $\tfrac{p}{2}$ where $\gcd{(p,2)}=1$. Thus $\lfloor a_k \rfloor a_k = a_k-\frac12$. By simple induction we can then get \[ a_{k+1}=a_k \cdot \left(a_0^{k+1}-\frac12 a_0^k-\frac12 a_0^{k-1}-\dots-\frac12 \right). \]Let \[ b_k = a_0^k-\frac12 a_0^{k-1}-\frac12 a_0^{k-2}-\dots-\frac12 \quad \text{for } k \geq 1. \]It then suffices to show that there is a number in the sequence $(b_k)_{k \in \mathbb{N}}$ that is even to get the desired contradiction, as it then follows that $a_k$ is an integer. Again, assume the contrary, in specific that all number $b_1,b_2,b_3,\dots$ are uneven. Then $b_{n+1}-b_n$ is even for all positive integers $n$. Now note \begin{align*} b_{n+1}-b_n &= a_0^{n+1}-\frac32 a_0^n \\ &= \left(\frac{2M+1}{2} \right)^{n+1}-\frac32 \left(\frac{2M+1}{2} \right)^{n} \\&= \left(\frac12 \right)^{n+1} \left((2M+1)^{n+1}-3 \cdot (2M+1)^n \right)\end{align*}It thus suffices to show that there is a $n$ such that \[ v_2 \left((2M+1)^{n+1}-3 \cdot (2M+1)^n \right) = n+1 \]as then $b_{n+1}-b_n$ would be uneven. But note \[ v_2 \left((2M+1)^{n+1}-3 \cdot (2M+1)^n \right) = v_2 \left((2M+1)^{n}(M-1) \cdot 2 \right) = v_2 (M-1) +1 \]For $M \geq 2$ the term $v_2(M-1)$ takes an positive integral value. Thus, we can choose $n := v_2(M-1)$ and will hence have found the desired contradiction. With that, we're done and indeed all $M \geq 2$ are solutions.

This was also Uzbekistan TST P2.

Answer:all $M>1$. If $M$=1,then $a_i$=$\frac{3}{2}$ $\forall$ $i$$\in$$\mathbb{N}$$_0$. If $M$ is even,then $a_1$=$a_0$$\lfloor$$a_0$$\rfloor$=($M$+$\frac{1}{2}$)$\lfloor$$M$+$\frac{1}{2}$$\rfloor$=$M^2$+$\frac{M}{2}$ is an integer. $Lemma1$:If $a_i$ is such that $\lfloor$$a_i$$\rfloor$ is even,then $a_{i+1}$ $\in$ $\mathbb{Z}$. $Proof$:$a_{i+1}=a_i$$\lfloor$$a_i$$\rfloor$=$a_{i-1}$$\lfloor$$a_{i-1}$$\rfloor$$\lfloor$$a_i$$\rfloor$=$\dots$=$a_0$$\lfloor$$a_0$$\rfloor$$\dots$$\lfloor$$a_i$$\rfloor$,$\lfloor$$a_i$$\rfloor$$\in$$\mathbb{Z}$$a_0$$\lfloor$$a_i$$\rfloor$=($2M+1$)$\frac{1}{2}$$\lfloor$$a_i$$\rfloor$ is an integer,hence $a_{i+1}$$\in$$\mathbb{Z}$. Now assume that $M$ is odd,$M$=$2^lk+1$,where $l$$\in$$\mathbb{N}$ and $k$ is an odd possitive integer. $Lemma2$:If $a_i$=$2^x$$k$+1+$\frac{1}{2}$,where $k$$\in$$\mathbb{N}$ is odd and $x$$\in$$\mathbb{N}$ then $a_{i+1}$=$2^{x-1}m+1+\frac{1}{2}$ ($m$ is an odd integer). $Proof$:$a_{i+1}$=$a_i$$\lfloor$$a_i$$\rfloor$=($2^x$$k$$+1+$$\frac{1}{2}$)($2^x$$k$$+1$)=$2^{x-1}k(2^{x+1}k+2^2+1$)+1+$\frac{1}{2}$=$2^{x-1}m+1+$$\frac{1}{2}$. Now $a_0$=$M$+$\frac{1}{2}$=$2^lk+1+$$\frac{1}{2}$.By $Lemma2$ $a_1$=$2^{l-1}y+1+$$\frac{1}{2}$ $\Rightarrow$ $a_2$=$2^{l-2}z+1+$$\frac{1}{2}$ $\Rightarrow$$\dots$ $\Rightarrow$ $a_l$=$2^{0}t+1+$$\frac{1}{2}$ $\Leftrightarrow$ $\lfloor$$a_l$$\rfloor$=$k+1$ is even (since $k$ is odd),hence by $Lemma1$ $a_{l+1}$ is an integer.

Another proof (because, why not?) Consider the sequence $\{b_i\}$, such that $a_i = \frac{b_i}{2}$ for all non-negative integers $i$. We prove the following lemma using induction: Lemma: If ${a_i}$ is not an integer for any non-negative integer $i$, then for all natural numbers $n$ $$b_i \equiv 3 mod 2^n$$ Proof: Base case: $n = 1$ is easy to see. Induction step: Suppose our lemma is true for $n = j$; $j \geq 1$ We'll prove it for $n = j + 1$ Using our induction hypothesis, for any non-negative integer $i$: $b_i = 2^j \cdot k + 3$ for some integer $k$ Then, $b_{i+1} = (2^j \cdot k + 3)(2^{j-1} \cdot k + 1)$ Using $b_{i+1} \equiv 3 mod 2^j$, we get $k = 2l$ for some integer $l$ This implies $b_i \equiv 2^{j+1} \cdot l + 3 \equiv 3 mod 2^{n+1}$, as required. Using the above lemma, if no $a_i$ is an integer, then for all natural numbers $n$: $$(2M + 1) \equiv 3 mod 2^n$$ This is possible iff $(2M + 1) - 3 = 0$, i.e. $M = 1$ It's easy to see that no $a_i$ is an integer for $M = 1$ Therefore, the answer is all natural numbers greater than $1$.

Sorry I don't know latex Here is my solution it is beautiful. {} Denotes floor function Assume that Ax is the required no. Then we have Ax=A(x-1)*{A(x-1)} And A(x-1)= A(x-2){A(x-2)} and similarly we get Ax=A0*({A1}*{A2}....{Ax+1}) since Ax is integer we want any of the term {Ak} to be even so it can cancel out with the 2 in the denominator of (2m+1)/2 which can be shown easily.

The answer is all $M>1$. Suppose for the sake of contradiction that for a certain $M>1$ the sequence doesn't contain any integer. Note that $b_k\stackrel{\text{def}}{=} \lfloor{a_k}\rfloor=a_k-\frac{1}{2}$, and also $b_k$ must be an odd integer for any $k$, otherwise $a_{k+1}$ would be integer, thus let $b_k=2c_k+1$. $$a_{k+1}=b_{k+1}+\frac{1}{2}=b_k\left(b_k+\frac{1}{2}\right) \implies 2c_{k+1}+1=(2c_k+1)^2+c_k \implies 2c_{k+1}=4c_k^2+5c_k$$This means that $c_k$ is even for any $k$, hence a standart infinite descent argument implies that $c_k=0$ for any $k$, but this holds only when $M=1$ and we reach a contradiction.

The answer is indeed all M > 1! Assume that the sequence doesn't contain any integers. Then for all k, $ \lfloor{a_k}\rfloor = 2n+1 $ for some positive integer n. Since this implies that $a_k$ is between [2n+1, 2n+2), $a_k=2n+1+\frac{1}{2}$. Applying this to the recursion, $a_{k+1}=4n^{2}+5n+1+\frac{1}{2}$. Now, we wish that $4n^{2}+5n+1$ is odd as we know that $a_{k+2}$ is not an integer. Thus, n is even. Now, $ \lfloor{a_k}\rfloor = 4m+1 $ for some m which implies that $a_{k+1}-\frac{1}{2}$ or $4n^{2}+5n+1$ has to be 1 in respect to (mod 4) for $a_{k+3}$ to be an integer. Thus, n is a multiple of 8. Applying the process again, n has to be a multiple of 16, then 32 and so on. This is because when trying to prove that n is a multiple of $2^{l}$ for some l, we already know that n is a multiple of $2^{l-1}$. Thus, in the expression $4n^{2}+5n+1$, we wish that $5n$ is 0 in respect to (mod $2^{l}$), which is analogous to saying that n is a multiple of $2^{l}$.

Solution: Suppose the sequence $a$ does not any integer term. Therefore, $\lfloor a_n \rfloor$ is odd and $\{a_n\} = \frac{1}{2}$ for all $n$. Let $a_n = 2b_n - \frac{1}{2}$, we have \[2b_{n+1} - \frac{1}{2} = \left(2b_n -\frac{1}{2}\right)\cdot (2b_n-1) = 4b_n^2-3b_n+\frac{1}{2}\implies 2b_{n+1} = 4b_n^2-3b_n+1.\]Notice that $b_n$ is integer, so $b_n$ is also odd for all $n$. Suppose $b_n\equiv 1 \pmod {2^k}$ for all $n$ for some $k\geq 1$. Define $b_n = c_n\cdot 2^k + 1$. We have, \[c_{n+1}\cdot 2^{k+1} + 2 = 4(c_n\cdot 2^k + 1)^2 - 3(c_n\cdot 2^k + 1) + 1 = c_n^2\cdot 2^{2k+2} + c_n\cdot 2^{k+3} + 4 - 3c_n\cdot 2^k - 2.\]Therefore we have, \[0 \equiv c_n^2\cdot 2^{2k+2}+c_n\cdot 2^{k+3} - 3c_n\cdot 2^k \equiv -3c_n\cdot 2^k\pmod {2^{k+1}}\implies 2\mid c_n ~~\forall~ n.\]This means by induction, $b_n\equiv 1 \pmod {2^k}$ for all $k$ and $n$ or $b_n = 1$ which gives $M = a_0-\frac{1}{2} = 2b_0 - \frac{1}{2}-\frac{1}{2} = 1$. All $M>1$ satisfy the condition. $~~\square$

Here is a clean solution: The answer is all $\boxed{M>1}$. For $M=1$, we see that $a_n=3/2$ always. Suppose we had $a_n\not\in\mathbb{Z}$ for all $n\ge 0$. We'll show $M=1$. Let $b_n=2a_n$, and note that $b_n$ must now always be an odd integer. Thus, \[b_{n+1}=b_n\lfloor b_n/2\rfloor=b_n\cdot\frac{b_{n-1}-1}{2}=\binom{b_n}{2}.\]We have the following lemma. Lemma: If $\binom{X}{2}\equiv 3\pmod{2^n}$ and $X$ is odd, then $X\equiv 3\pmod{2^{n+1}}$. Proof of Lemma: We see that $X(X-1)\equiv 6\pmod{2^{n+1}}$, so $2^{n+1}\mid (X-3)(X+2)$. But $X+2$ is odd, so $2^{n+1}\mid X-3$, as desired. $\blacksquare$ We have $b_N\equiv 3\pmod{2}$ for any $N\ge 0$, so by the lemma, $b_{N-1}\equiv 3\pmod{4}$, so again by the lemma $b_{N-2}\equiv 3\pmod{8}$. Eventually, we get \[b_0\equiv 3\pmod{2^{N+1}}.\]This is true for all $N\ge 0$, so we must have $b_0=3$, or $M=1$. $\blacksquare$

I liked this problem a lot. It was simple yet elegant . Here's my solution: Obviously $M=1$ doesn't work. We claim that there exists an integer term for all $M>1$. FTSOC assume that some $M>1$ does not satisfy our claim. By an easy induction, one can show that, for this to happen, we must have $a_i=\frac{1}{2}(2x_i+1)$ for some positive integer $x_i$. Then, we get that $$\frac{1}{2}(2x_{i+1}+1)=a_{i+1}=a_i \lfloor a_i \rfloor=\frac{2x_i+1}{2} \times x_i \Rightarrow 2x_{i+1}+1=x_i(2x_i+1) \Rightarrow x_{i+1}=x_i^2+\frac{x_i-1}{2}$$As $x_{i+1}$ and $x_i$ are both integers, so we get that $2 \mid x_i-1$, i.e. $x_i$ is odd for all $i \in \mathbb{N}$. Let $x_i=2y_i+1$ for $y_i \in \mathbb{N} \cup \{0\}$. Then, one easily finds that $$2y_{i+1}+1=(4y_i^2+4y_i+1)+y_i \Rightarrow 2y_{i+1}=y_i(4y_i+5)$$Thus, $y_i$ is even, and so we can take $y_i=2z_i$. This enables us to find $4z_{i+1}=2z_i(8z_i+5)$, which gives that $z_i$ is also even. Continuing in a similar fashion, we have $2^k \mid y_i$ for all $k \in \mathbb{N}$, which is only possible if $y_i=0 \text{ } \forall i \in \mathbb{N}$. This means that $x_i=1$, and $a_i=\frac{3}{2}$ for all positive integral values of $i$. However, this is never possible for $a_0 \geq \frac{5}{2}$. Hence, done. $\blacksquare$

We claim that every $M \geq 2$ is a solution note how $M=1$ doesn't work, as $a_i = a_j \forall i,j$. also notice how it each non-integer term in the sequence has $a_k - \lfloor a_k \rfloor = \frac{1}{2}$, hence it suffices to show that the floor of one of the terms is an even number, as that would imply that the next number would be an integer claim: $2M \equiv 0 \mod{4}$ works proof: that statement means that $M$ itself is an even integer, therefore $\lfloor M + \frac{1}{2} \rfloor = M$ is also an even integer, meaning $a_1 \in \mathbb{Z}$ . claim: apart from $2$, $2M \equiv 2 \mod{4}$ also works proof: In this case, M is an odd number. Notice how $a_{k+1} = \lfloor a_k \rfloor ^2 + \frac{\lfloor a_k \rfloor}{2}$. We may assume that $\lfloor a_k \rfloor \equiv 1 \mod{2}$ as else we would be done with $a_{k+1}$, therefore $\lfloor a_k \rfloor ^2$ is an odd number. We now split into two cases: c1: $\lfloor a_k \rfloor \equiv 3 \mod{4}.$ this means that $\lfloor \frac{\lfloor a_k \rfloor}{2}\rfloor \equiv 1 \mod{2}.$ However, $a_{k+1} = \lfloor a_k \rfloor (\lfloor a_k \rfloor + \frac{1}{2}) = \lfloor a_k \rfloor ^2 + \frac{\lfloor a_k \rfloor}{2}$ But for $a_{k+2}$ we take the floor of this expression which is the same as taking the floor of the last part, which results in an odd + odd case, which is even, which means that $a_{k+1}$ is an integer c.2: $\lfloor a_k \rfloor \equiv 1 \mod{4}$, then $\lfloor \frac{\lfloor a_k \rfloor}{2}\rfloor \equiv 0 \mod{2}$. claim: $v_2(\lfloor a_k \rfloor -1) > v_2(\lfloor a_{k+1} \rfloor -1)$ This would imply that at some point, we would hit c.1 However, this is clearly the case as $\lfloor a_{k+1} \rfloor= \lfloor a_k \rfloor ^2 + \lfloor \frac{\lfloor a_k \rfloor}{2}\rfloor$ is just always dragging a power of 2 out of the second part, which means that is at some point has to be an odd number, in which case the entire thing will be even Hence, we are done

We show that all $M>1$ work. Clearly M=1 doesn't work since the sequence stays at 3/2. Now, we induct on $n=v_2(a_n-1)$, noting that we want to prove it eventually becomes 0, from which it would follow that $a_n$ is even; this would then imply that floor[$a_k$] is even while $v_2{a_k}$ is always at least -1 (since it starts with at least -1, and each iteration does not divide by 2). Suppose for some M=k it reduces to 0. The next term is $$\frac{2M^2+M}{2}=\frac{2M^2+M-1}{2}+\frac 12\implies v_2(\frac{2M^2+M-1}{2}-1)=v_2((2M+3)(M-1))-1=v_2(M-1),$$so it reduces to the inductive hypothesis, as desired. $\blacksquare$

We, claim, that all $M>1,$ work. Note, that $M=1,$ doesn't work, because, the sequence, stays constant at $\frac{3}{2}.$ Case 1: $M,$ is even. If, $M,$ is even, we have that $a_1=M(M+\frac{1}{2}),$ which is an integer. Case 2: $M,$ is odd. Now, $a_1=\frac{2M+M-1}{2}+\frac{1}{2},$ so $v_2(a_1)=v_2((2M+3)(M-1))-1=v_2(M-1),$ and now we are done. (note we subtracted 1 because of the denominator)

We claim the answer is $\boxed{M>1}$. Clearly $M=1$ does not have an integer term as every term is $\frac{3}{2}$. Now, consider $\nu_2(M-1)$. We would eventually want it to equal $0$ since that would be equivalent to $M$ being even, which is a vacuous case. If $\nu_2(M-1) \neq 0$, \[a_1 = M \left(M+\frac{1}{2} \right) = \frac{2M^2+M-1}{2} + \frac{1}{2}.\] Letting $\frac{2M^2+M-1}{2}=N$, we have that \[\nu_2(N-1) = \nu_2\left(\frac{2M^2+M-3}{2}\right) = \nu_2((2M+3)(M-1))-1 = \nu_2(M-1)-1.\] Thus, we can repeat this process until the value equals $0$, at which point the term will be an integer.

All integers $M>1$ work. If $M$ is even, $a_1 \in \mathbb Z$. Otherwise assume $M=2^k\cdot m + 1$ where $k\ge 1$ is maximal and so $m$ is odd. I claim that $a_{k+1} \in \mathbb{Z}$. Proof follows by using induction on $k$. For $k=0$, $M=m+1$ where $m$ is odd. Thus $a_0 = m+1 + \dfrac{1}{2}$, $a_1 = \left(m+1 + \frac{1}{2}\right)(m+1) = (m+1)^2 +\frac{m+1}{2}$ which is an integer as $m$ is odd. Now we assume for $k$ and prove it for $k+1$. We have $a_0 = 2^{k+1}m + 1 + \frac{1}{2}$, $a_1 = \left(2^{k+1}m + 1 + \frac{1}{2}\right)(2^{k+1}m+1) = 2^{2(k+1)}m^2 + 2^{k+2}m + 1 + 2^k m + \frac{1}{2} = 2^{k+2}(2^{k}m^2 + m) + \left(2^km + 1 + \frac{1}{2}\right)$. Firstly note that $2^{k+2}(2^{k}m^2 + m)$ is an integer itself. Now note that after each move, the power of two in $2^{k+2}(2^{k-1}m^2 + m)$ decreases at most by $1$ due to the $2$ in the denominator after multiplication. So after $k+1$ moves, the power of $2$ remaining in it is at least $2^1$. But note that after $k+1$ moves, the $2^km + 1 + \frac{1}{2}$ turns into an integer due to induction hypothesis. Thus we are done.

We claim the the answer is $M \ge 2$. It is clear that $M = 1$ is invalid, as all terms are $\frac{3}{2}$ and thus will never become integer. For the sake of contradiction, assume for some $M \ge 2$, we have $a_k = b_k + \frac{1}{2}$ for all nonnegative integers $k$, such that $b_k$ is an integer. Then all $b_k$ are odd, otherwise an integer term would exist. Then $$b_{k + 1} + \frac{1}{2} = b_k \left(b_k + \frac{1}{2} \right) \implies 2b_{k + 1} = 2b_k^2 + b_k - 1.$$Now taking the difference from $k = t, t + 1$ yields $$2b_{t + 2} - 2b_{t + 1} = (b_{t + 1} - b_t)(2b_{t + 1} + 2b_t + 1).$$Since $M \ne 1$, all $b_i$ are distinct, so $2 \cdot \frac{b_{t + 2} - b_{t + 1}}{b_{t + 1} - b_t}$ is an odd integer for all nonnegative integers $t$, implying that $\nu_2(b_{k + 1} - b_k)$ decreases as we progress through the sequence. At some point, we reach a sufficient $k = m$ where $b_{m + 1} - b_m$ must be odd, a contradiction as all $b_k$ are odd. Our proof is complete.

The answer is $M \ge 2.$ The first thing is to prove that $M = 1$ doesn't work. This is clear, since $a_0 = \frac{3}{2},$ so $a_1 = 1 \cdot \frac{3}{2} = \frac{3}{2}$ and the sequence is always a constant non-integer. Now we show that all other $M$ work. Claim: Write $M = b \cdot 2^c + 1$ for $b$ odd. Then $a_{c+1}$ is an integer. Proof: We induct on $c.$ The base case $c = 0$ is obvious since $M$ is even and thus $M(M+0.5) = a_1$ is an integer. Now suppose that the claim holds for $c = k-1;$ we will use it to show that the claim holds for $c = k.$ We start with the number $a_0 = M + \frac{1}{2} = b \cdot 2^c + \frac{3}{2}.$ Then \begin{align*} a_1 &= \left(b \cdot 2^c + \frac{3}{2}\right)(b \cdot 2^c + 1) \\ &= b^2 \cdot 2^{2c} + b \cdot 2^c + 3b \cdot 2^{c-1} + \frac{3}{2} \\ &= (b^2 \cdot 2^{c+1} + 2b + 3b) \cdot 2^{c-1} + \frac{3}{2} \\ &= (b^2 \cdot 2^{c+1} + 5b) \cdot 2^{c-1} + 1 + \frac{1}{2}. \end{align*}Note that $b^2 \cdot 2^{c+1} + 5b$ is odd since $b$ is odd. Applying the inductive hypothesis on $a_1$ as the starting term gives us that $a_{1+k}$ is an integer. The induction is complete. $\square$ Therefore, $a_{c+1}$ is an integer, so all $M \ge 2$ work. These are the only solutions, and we are done.

Claim: if M-1 has v_2 of n, then we win in n+2 steps. For example if M is 9 mod 16 then M-1 has v_2 of 3, and after one step we go to a number with integer part 5 mod 8, then a number with integer part 3 mod 4, then a number with even integer part, then we finally get to an integer. Note that the union of numbers cong to 0 mod 2, numbers congruent to 3 mod 4, 5 mod 8, etc etc give you every number except 1, by CRT. The proof is trivial induction. The base cases for 3 mod 4, even integer part, integer are trivial. Thus we want to prove that if M = 2^n + 1 (mod 2^(n+1)), then the next element in the sequence has integer part M_1 = 2^(n-1) + 1 (mod 2^(n)), Expanding (M)(M+0.5) in mod 2^(n+1) gets us 2^(2n)+2^(n+1)+1 + 2^(n-1) + 0.5, which reduces to 1 + 2^(n-1) + 0.5 (mod 2^(n+1), and all numbers like this have integer part 1 + 2^(n-1) + k*2^(n+1), and taking this mod 2^n finishes the induction. Trivial problem. Thus we have a construction for all numbers except 1, which gets stuck at 1.5 forever.

We have that $a_0 = M + \frac{1}{2} = \frac{2M + 1}{2}$ $\Rightarrow$ if M is even we have that $a_1 = \frac{2M + 1}{2}\cdot M$ and since $2 \mid M$, we have that $a_1$ is an integer. So obviously every even M works. Now assume M is odd. For M = 1, it doesn't work. Now we will prove by induction that each M works. Claim: For all $M > 1$, we will always get an integer. We will now induct on $\nu_2(M-1)$. If it is 0, then it becomes an integer in one step and if not, then the next term will be $\frac{2M^2+M}{2} = \frac{2M^2+M-1}{2}+\frac {1}{2}$ $\Rightarrow$ if $\frac{2M^2+M-1}{2} = K$, we have that $\nu_2(K-1) = \nu_2((M-1)(2M-3)/2) = \nu_2(M-1)-1$, so the value reduces and the inductive hypothesis will end.

We claim it is all $\boxed{M>1}$. Clearly $M=1$ fails because $a_i=\frac32$ for all $i$. Clearly all even $M$ work because $a_1=M^2+\frac{M}{2}$. Therefore, assume $M$ odd. Since $M$ is odd, we can let $M=2k+1$. Then, $a_1=(2k+\frac32)(2k+1)=4k^2+5k+\frac32=(k+1)(4k+1)+\frac12$. Since $(k+1)(4k+1)$ is even when $k$ is odd, an odd $M$ works if $k=\frac{M-1}2$ is odd. Otherwise, $k$ is not odd, so let $k=2j$. Then $a_1=(2j+1)(8j+1)+\frac12$. Since $\frac{(2j+1)(8j+1)-1}{2}=8j^2+5j$ is odd when $j$ is odd, an odd $M$ works if $j=\frac{M-1}{4}$ is odd. Otherwise, $j$ is not odd, so let $k=4j$. Then, $a_1=(4j+1)(16j+1)+\frac12$. Since $\frac{(4j+1)(16j+1)-1}4=16j^2+5j$ is odd when $j$ is odd, an odd $M$ works if $j=\frac{M-1}{8}$ is odd. Let $k=2^nj$ and assume that an odd $M$ works when $\frac{M-1}{2^n}$ is odd for $n\ge 3$. Then, $a_1=(2^nj+1)(2^{n+2}j+1)+\frac12$. Since $\frac{(2^nj+1)(2^{n+2}j+1)-1}{2^n}=2^{n+2}j^2+5j$ is odd when $j$ is odd, an odd $M$ works if $j=\frac{M-1}{2^{n+1}}$ is odd. By induction, an odd $M$ works when $\frac{M-1}{2^a}$ is odd for some integer $a\ge 1$. This is clearly true for all odd $M>1$, as desired.

The answer is all $M > 1$. To see that $M = 1$ fails, note that $a_0 = \frac 32$ and $a_1 = \frac 32$, so the sequence is constant. Now we show that $M > 1$ works. The denominator of the sequence is always either $2$ or $1$ (trivial by induction).Now we rephrase the problem in terms of the number on the numerator while the denominator is constant, consider the sequence $b$ where $b_0 = 2M + 1$. Then if $b_i \equiv 1 \mod 4$, $i$ is the last element of the sequence, otherwise we have $a_{i + 1} = \frac{b_i}{2} (\frac{b_i - 1}{2})= \frac{b^2_i - b_i}{4}$, so $b_{i + 1} = \frac 12 (b^2_i - b_i)$. Now we show that $\nu_2 (b_i - 3)$ decreases by $1$ for each increment in $i$. At some point this forces $b_i \equiv 1 \mod 4$ and we are done. Let $\nu_2 (b_i - 3) = c$, then we have $b_i \equiv 2^{c} + 3 \mod 2^{c + 1}$, so we know $b_{i + 1} = (2^cx + 2^{c -1} + 1)(2^{c + 1}x + 2^c + 3) = 2^{c} (\mathrm{stuff}) + 3*2^{c - 1} +3$ which is $2^{c - 1} + 3 \mod 2^{c}$, so $\nu_2 (b_{i + 1} - 3) = c - 1$ as desired.

The answer is all $M>1$. Clearly $M=1$ does not satisfy the desired condition, and $a_1$ is an integer for all even $M$. Suppose that $a_k=N+\frac12$ for any index $k$ and odd $N\ge3$. Then $a_{k+1}=\frac{(N-1)(2N+3)}2+\frac32$, so that $\nu_2(a_{k+1}-\frac32)=\nu_2(a_k-\frac32)-1$. Hence, for some large index $k'$ we have $\nu_2(a_{k'}-\frac32)=0$, implying that $a_{{k'}+1}$ is an integer. Setting $(M,0)=(N,k)$ for any odd $M\ge3$ finishes. $\blacksquare$

Originally solved as part of Rohan Goyal's "Sequences" Lecture at the 2024 India IMO Training Camp. We find all $M$ such that no term in the sequence is an integer. Indeed, we claim that $2^n \mid M-1$ for all $n$. The proof is by induction. Obviously $M$ is odd, else $a_1 \in \mathbb{Z}$. Now assume that $2^n \mid M-1$. Then note that shifting the sequence as $a_{k+1} \to a_k$, we must have $\lfloor a_1 \rfloor \equiv 1 \pmod{2^n}$. As it turns out, this is nothing but $$\frac{2M^2 + M-1}{2} \equiv 1 \pmod{2^n} \implies 2^{n+1} \mid (2M+3)(M-1) \implies 2^{n+1} \mid M-1$ as desired, But now $M-1$ has infinite $\nu_2$, so $M = 1$ is the only value for which no term is an integer. And indeed no term is an integer then as the sequence consists wholely of $\frac 32$. Thus the final answer is $M \in \mathbb{N} \setminus \{1\}$.

The answer is all $\boxed{M > 1}$. $M=1$ obviously does not work. The idea is casework on $v_2(M-1)$. If $M-1 = b \cdot 2^c$ for $b$ odd, then I claim $a_{c+1}$ is the desired integer. The most straightforward way to do so is induction. The $c=0$ case is immediate from $a_1 = M^2 + \frac{M}{2}$. If we suppose that it is true for $c=k$, we wish to prove it is so for $c=k+1$. For $M = b \cdot 2^{k+1}+1$, then $a_1 = (b \cdot 2^{c+1}+1)^2 + b \cdot 2^c + \frac12$ which when viewing $M ' = (b \cdot 2^{c+1}+1)^2 + b \cdot 2^c$ has $v_2(M'-1) = c$. Using our inductive hypothesis finishes.

Clearly if $M$ is even $a_1$ is an integer. If $M \equiv 3 \pmod 4,$ then write $M=4m-1,$ then $$a_1=(4m-1)(4m-1/2)=16m^2-6m+1/2,$$which has an even integer part so $a_2$ is an integer. Now, suppose that $M \equiv 1 \pmod 4,$ and $M > 1.$ For the sake of a contradiction, assume that none of the terms will be integers. Then clearly all $\lfloor a_k \rfloor \equiv 1 \pmod 4.$ Write $M=4m+1,$ we have that $$a_1=(4m+1)(4m+3/2) = 16m^2+10m+3/2.$$Hence $m$ is even. Let $m=2m_0.$ Then we can write $$a_1=4(16m_0^2+5m_0)+1+1/2.$$Similar to above, for $\lfloor a_2 \rfloor \equiv 1 \pmod 4$ we must have that $m_0$ is even, so continuing this logic eventually yields a contradiction as $v_2(M-1)$ is finite. Thus we have shown that all positive integers $M \geq 2$ work. Clearly, $M=1$ does not work as $a_k = \frac32$ for all $k,$ so $M \geq 2$ are the only solutions.