Consecutive semiprimes

4 is a semiprime, but 3 and 5 are not.

Suppose there was a list of 4 consecutive semiprimes. One of them would be a multiple of 4. Say n*4. But this multiple of 4 cannot be a semiprime for n>1 because it has at least three factors. Therefore the answer is No, there are no strings longer than three.

Simple,

Let suppose those numbers are $a$ , $a+1$ , $a+2$ , $a+3$ .

Start: Except 4 itself, any number that is divisible by 4 will be of the form 4xm i.e. 2x2xm. It means it has 'at least' 3 primes.

Case 1: Dividing $a$ by 4 gives remainder 0.

Case 2: Dividing $a$ by 4 gives remainder 1.

Case 3: Dividing $a$ by 4 gives remainder 2.

Case 4: Dividing $a$ by 4 gives remainder 3.

Explained: In case 1: $a$ is divisible by 4; case 2: $a+3$ is divisible by 4; case 3: $a+2$ is divisible by 4; and case 4: $a+1$ is divisible by 4.

Hence any sequence of 4 consecutive numbers would have a number that is divisible by 4 (two 2's) and at least one other prime. (e.g. 87,88,89,90: 88 is divisible by 2, 2, and 11 and one more 2).

Concluded: Hence we don't have any sequence of 4 or more numbers with all of them being semiprime.

:)

A string longer than 3 means there are even multiples of two consecutive numbers. As two consecutive numbers cannot be prime, a string longer than 3 doesn't exist

EDIT: One exception to my solution is 2 and 3, which are the only two consecutive primes. But that string would be for something from 4 to 6, and as they are not biprimes, we can safely conclude our result

Any list of 4 consecutive numbers will include a number divisible by 4. Any number divisible by 4 is also divisible by 2 so it is not a semi-prime.

If there was a run of four semiprimes then two of them would be consecutive evens. Call these 2n and 2(n+1). Either n or n+1 has to be even. So one of the semiprimes would have to be 2 times an even number, which means is contains another factor of 2, which means it is not semiprime.

So, by contradiction, there can be no runs of semiprimes longer than three.

• A list of more than 3 consecutive semiprimes contains at least 2 even numbers.
• To get an even semiprime, you need to multiply a prime times 2, since you can only get an even number from multiplication if one of the factors is even and 2 is the only even prime.
• Let the first even semiprime be $\boxed{prime_1 * 2 = n}$ and the next one $\boxed{prime_2 * 2 = n + 2}$
• From this we can see that the distance between those primes needs to be 1. Because all primes except 2 are odd and adding 1 to any odd number results in an even one, we get that $prime_2$ needs to be even and so, cannot be prime. Therefore, no list of more than 3 consecutive semiprimes is possible.

In any four consecutive numbers, there will be two numbers which are even and differ by $2$ . Since they are semi-primes, they are of the form $2p$ and $2(p+1)$ , where $p$ and $p+1$ are primes. Clearly, the only choice for $p$ is $p=2$ . However, none of the possible sequences $\{4,5,6,7\}$ or, $\{3,4,5,6\}$ constitute a run of semiprimes.

Let's check if we can do four consecutive semiprimes. Any consecutive list of four numbers will contain two evens and two odds. The two evens will be $2 \times x$ and $2 \times (x+1)$ , where $x$ and $x+1$ are primes. We see that two consecutive primes ( $x$ and $x+1$ ) are needed: the only consecutive primes are $2$ and $3$ . Plugging in, we get the lists $(4, 5, 6, 7)$ and $(3, 4, 5, 6)$ , but $3$ , $5$ , and $7$ are all primes, not semiprimes. So the answer is $\boxed {NO}$

Here is a proof by contradiction.

Suppose there is a run of four consecutive semi-primes.

Then it contains two consecutive even numbers. Let's call them 2n and 2(n+1).

Since these numbers are in the run, they are semi primes, which means that n and (n+1) are primes.

The only two consecutive integers which are prime are 2 and 3, so n=2.

So the two even numbers in the run are 4 and 6.

The number 5 which is sandwiched between them is also in the run.

But 5 is not a semiprime!

Can we have a run of four semiprimes, one of which is not a semiprime??

Contradiction!

So we conclude that the original bold supposition is false.

And so $\boxed{\text{there is no run of four semiprimes}}$

If there are four consecutive semiprime numbers, the sequence has to be: Even/odd/even/odd or odd/even/odd/even Beyond number two, there is no other even prime number. And i can not reach an even one as product of two odds. So, due to the fact that there is only one even prime number, there are no runs longer than three numbers.

For any consecutive numbers with length more than 4, there $must$ have at least a number is multiple of $4$ .

If these consecutive semiprimes have 4 in it, noted that $3$ and $5$ is a prime, therefore consecutive semiprimes with $4$ in it has only length of 1.

If these consecutive numbers contains other multiple of $4$ (let us say that one of them is $m$ ) , then $m=4n=2^2 \times n$ . Since $n \neq 1$ , therefore $m$ is not a semiprime.

Conclusion: there are no run of consecutive semiprimes longer than three numbers.

For each 3 consecutive semiprimes, the 1st and 3rd number have to be odd and the 2nd number has to be even (with a single 2 factor times a prime). If we observe the 4th number it will always have another 2 factor which will prevent it of being a prime.

There's no such run. Let's consider the following sequence: n , n+1 , n+2 , n+3 . In this sequence, we must have two even numbers such that the difference betwenn then is 2. Let's say that n and n+2 are this numbers. So: n=2q n+2=2q' , where q and q' are both primes. This lead us to 2q=2q'-2 , and hence q'=q+1 , which only holds true if q=2 , because any prime greater than two would be odd, hence q+1 would be even and can't be a prime.

One of the four must be a multiple of 4 & that multiple won't be a semiprime. So no longer strings than 3

We know that since a multiple of $2$ occurs every two consecutive integers, therefore either $n$ or $n+1$ is a multiple of $2$ .

Since we are talking about consecutive integers if we let $n + 1 = 2\times p_1$ where $p_1$ is prime other than two, then the next multiple of two will be $n + 3 = 2\times (p_1 + 1)$ . But $p_1 + 1$ is even and therefore we can write it as $p_1 = 2\times p_2$ which would mean that $n + 3$ is not semiprime. This also means that if we want a set of 3 consecutive semiprimes, our starting value in the chain $n$ cannot be even.

As a result we can say that the maximum length for consecutive numbers which are semiprime is $3$ .