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P(4,1) is true because of 2,3,4,5 but I strongly doubt that P(5,1) is true. Indeed P(4,1) is only true that once and even P(3,1) happens for the last time at 7,8,9. Of any 210 consecutive integers one is divisible by 2,3,5,7 so P(210,3) fails. I'm sure that can be improved. It does generalize .

As noted in the comments, there are two cases (at least) of 8 consecutive integers each with two distinct prime factors and 12 can be ruled out. Actually it seems extremely unlikely that there are any cases of 9, let alone 10 or 11: Such a run of 9 would have to include 7 integers $3(a-1),2(b-1),*,2^i3^j,*,2(b+1),3(a+1)$ where $a\pm1=2^{i}3^{j-1}\pm1$ and $b\pm1=2^{i-1}3^{j}\pm1$ are two pairs of twin primes (or merely twin prime powers). I would expect only finitely many cases where $2^u3^v\pm1$ are both prime powers (I find 35 such up to $10^{50}$ with $u,v \ge 1$, let alone getting two such pairs as above ( I find 4 cases, 36,144,216 and $2^{33}3^9$.) This leaves out the requirements that we would also need $2^{i}3^{j}\pm1$ to each have two distinct prime factors (ruling out the smallest and largest) and have at least one of $2^{i}3^{j}\pm2$ 2^{i}3^{j}\pm4$to be 4 times a prime power, and more. 5 TeX P(4,1) is true because of 2,3,4,5 but I strongly doubt that P(5,1) is true. Indeed P(4,1) is only true that once and even P(3,1) happens for the last time at 7,8,9. Of any 210 consecutive integers one is divisible by 2,3,5,7 so P(210,3) fails. I'm sure that can be improved. It does generalize . As noted in the comments, there are two cases (at least) of 8 consecutive integers each with two distinct prime factors and 12 can be ruled out. Actually it seems extremely unlikely that there are any cases of 9, let alone 10 or 11: Such a run of 9 would have to include 7 integers $3(a-1),2(b-1),*,2^i3^j,*,2(b+1),3(a+1)$ where$a\pm1=2^{i}3^{j-1}\pm1$and$b\pm1=2^{i-1}3^{j}\pm1$are two pairs of twin primes (or merely twin prime powers). I would expect only finitely many cases where$2^u3^v\pm1$are both prime powers (I find 35 such up to$10^50$10^{50}$ with $u,v \ge 1$, let alone getting two such pairs as above ( I find 4 cases, 36,144,216 and $2^{33}3^9$.) This leaves out the requirements that we would also need $2^{i}3^{j}\pm1$ to each have two distinct prime factors (ruling out the smallest and largest) and have at least one of $2^{i}3^{j}\pm2$ to be 4 times a prime power, and more.

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P(4,1) is true because of 2,3,4,5 but I strongly doubt that P(5,1) is true. Indeed P(4,1) is only true that once and even P(3,1) happens for the last time at 7,8,9. Of any 210 consecutive integers one is divisible by 2,3,5,7 so P(210,3) fails. I'm sure that can be improved. It does generalize .

As noted in the comments, there are two cases (at least) of 8 consecutive integers each with two distinct prime factors and 12 can be ruled out. Actually it seems extremely unlikely that there are any cases of 9, let alone 10 or 11: Such a run of 9 would have to include 7 integers $3a,2b,*,2^i3^j,*,2(b+2),3(a+2)$3(a-1),2(b-1),*,2^i3^j,*,2(b+1),3(a+1)$ where$a\pm1=2^{i}3^{j-1}\pm1$and$b\pm1=2^{i-1}3^{j}\pm1$are two pairs of twin primes (or merely twin prime powers). I would expect only finitely many cases where$2^u3^v\pm1$are both prime powers (I find 35 such up to$10^50$with$u,v \ge 1$, let alone getting two such pairs as above ( I find 4 cases, 36,144,216 and$2^{33}3^9$.) This leaves out the requirements that we would also need$2^{i}3^{j}\pm1$to each have two distinct prime factors (ruling out the smallest and largest) and have at least one of$2^{i}3^{j}\pm2\$ to be 4 times a prime power, and more.

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