Today is the $88^{th}$ day of the year.

$88 = 2^3 \times 11$

$88$ is a member of two primitive Pythagorean triples

$(88, 105, 137)$

$(88, 1935, 1937)$

$88$ is an Erdos–Woods number which is defined as: there exists a positive integer $a$ such that in the sequence $(a, a + 1, …, a + k)$ of consecutive integers, each of the elements has a common factor, other than one, with one of the endpoints.

For example, consider $a = 2,184$ and $k = 16$ which gives us the set of $17$ numbers from $2,184$ to $2,200$ inclusive:

$2,184$ has a common factor of $2,184$ with $2,184$

$2,185$ has a common factor of $5$ with $2,200$

$2,186$ has a common factor of $2$ with $2,184$

$2,187$ has a common factor of $3$ with $2,184$

$2,188$ has a common factor of $4$ with $2,184$

$2,189$ has a common factor of $11$ with $2,200$

$2,190$ has a common factor of $6$ with $2,184$

$2,191$ has a common factor of $7$ with $2,184$

$2,192$ has a common factor of $8$ with $2,184$

$2,193$ has a common factor of $3$ with $2,184$

$2,194$ has a common factor of $2$ with $2,184$

$2,195$ has a common factor of $5$ with $2,200$

$2,196$ has a common factor of $4$ with $2,200$

$2,197$ has a common factor of $13$ with $2,184$

$2,198$ has a common factor of $2$ with $2,200$

$2,199$ has a common factor of $3$ with $2,184$

$2,200$ has a common factor of $2,200$ with $2,200$

So, there is a number $a$ such that every single one of the numbers in the set $(a, a + 1, …, a + 88)$ has a common factor greater than $1$ with one of $a$ or $a + 88$, see A059756.

I can neither find what this $a$ is nor how this sequence has been calculated (Note that the first $1,052$ members of the sequence can be found here). However, sequence A05975 does gives the first few values for $a$. Thus we learn that the ninth member has $k = 70$ and $a = 13,151,117,479,433,859,435,440$. So the thirteenth member with $k = 88$ must be an even larger number.

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