Assuming that $x$ and $y$ are positive integers:
At first I tried to see what will happen For $x^2<y<(x+1)^2$ where $x^2$ is the nearest perfect square to $y$, with the sequence $(y-1), (y-1-3),(y-1-3-5),...,(y-1-3-5-...-2x-1)$
*Note: $(1),(1+3),(1+3+5),(1+3+5+7)...$ are the square numbers.
In the following examples, I keep deducting until I either hit the first prime or until I hit the last positive result before becoming negative.
Also I don't include $y$ when it is a perfect square, because there can not be a perfect square between $x^2$ and $(x+1)^2$
Examples:
$y = 6$ : $2^2<6<3^2$ : $6-1=5$ Prime
$y = 7$ : $2^2<7<3^2$ : $7-1-3=3$ Prime
$y = 8$ : $2^2<8<3^2$ : $8-1-3=4$ Not Prime
$y = 9$ : $9$ is perfect square
$y = 10$ : $3^2<8<4^2$ : $10-1-3-5=1$ Not Prime
$y = 11$ : $3^2<11<4^2$ : $11-1-3=7$ Prime
$y = 12$ : $3^2<12<4^2$ : $12-1=11$ Prime
$y = 13$ : $3^2<13<4^2$ : $13-1-3-5=4$ Not Prime
...
$y = 31$ : $5^2<31<6^2$ : $31-1-3-5-7-9=6$ Not Prime
...
$y = 58$ : $7^2<31<8^2$ : $58-1-3-5-7-9-11-13=9$ Not Prime
...
Conclusion: Sometimes it will end in a prime and sometimes in a composite.
On the contrary to the above, in the following (so far):
For $x^2<y<(x+1)^2$ and where $y > 2$ and where $x^2$ is the nearest perfect square to $y$, it seems as if there will always be a at least one prime for the sequence $(y-2x-1), (y-2x-1-2x-3),(y-2x-1-2x-3-2x-5),...,(y-2x-1-2x-3-2x-5-,...,-2x-(2x-1))$.
*note: $2x-(2x-1)=1$.
Examples:
$y = 3$ : $1^1<3<2^2$ : $3-1=2$ Prime
$y = 4$ : $y$ is a perfect square
$y = 5$ : $2^2<5<3^2$ : $5-3=2$ Prime
$y = 6$ : $2^2<6<3^2$ : $6-3=3$ Prime
$y = 7$ : $2^2<7<3^2$ : $7-3-1=3$ Prime
$y = 8$ : $2^2<8<3^2$ : $8-3=5$ Prime
$y = 9$ : $y$ is a perfect square
$y = 10$ : $3^2<10<4^2$ : $10-5=5$ Prime
$y = 11$ : $3^2<11<4^2$ : $11-5-3=3$ Prime
$y = 12$ : $3^2<12<4^2$ : $12-5=7$ Prime
$y = 13$ : $3^2<13<4^2$ : $13-5-3=5$ Prime
$y = 14$ : $3^2<14<4^2$ : $14-5-3-1=5$ Prime
$y = 15$ : $3^2<15<4^2$ : $15-5-3=7$ Prime
$y = 16$ : $y$ is a perfect square
$y = 17$ : $4^2<17<5^2$ : $17-7-5=5$ Prime
$y = 18$ : $4^2<18<5^2$ : $18-7=11$ Prime
$y = 19$ : $4^2<19<5^2$ : $19-7-5=7$ Prime
$y = 20$ : $4^2<20<5^2$ : $20-7=13$ Prime
$y = 21$ : $4^2<21<5^2$ : $21-7-5-3-1=5$ Prime
$y = 22$ : $5^2<22<5^2$ : $22-7-5-3=7$ Prime
$y = 23$ : $5^2<22<5^2$ : $23-7-5=11$ Prime
$y = 24$ : $5^2<24<5^2$ : $24-7=17$ Prime
...
For $x^2<y<(x+1)^2$ where $x^2$ nearest to $y$ and where $y>2$, there seems to always be a prime in the sequence $(y-2x-1), (y-2x-1-2x-3),...,(y-2x-1-2x-3-...-2x-(2x-1))$?
- I didn't have enough space to add "and where $y>2$" in the title
Am I correct and if so, is it a flux or is there any explanation for why it will always result in at least one prime, and why in the former sequence example it does not always result in at least one prime?