Score 32 34 36

{5, 7, 11, 13, 17, 23, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 101, 113, 131, 137, 151, 211, 229, 241, 281, 313, 379, 401, 433, 457, 491, 521, 571, 601, 25117, 293362609}

This is an improvement of Arnauld's answer. I just noticed that

\$ \dfrac{1}{19-1}=\dfrac{1}{73-1}+\dfrac{1}{61-1}+\dfrac{1}{41-1} \$

But 41 and 61 were already used in Arnauld's answer so I had to then figure out that

\$ \dfrac{1}{61-1} = \dfrac{1}{151-1} + \dfrac{1}{101-1} \$ and \$ \dfrac{1}{41-1} = \dfrac{1}{281-1}+\dfrac{1}{211-1}+\dfrac{1}{151-1}+\dfrac{1}{101-1} \$

But now I am using 151 and 101 twice. So I spent some time and discovered that

\$ \dfrac{1}{151-1} = \dfrac{1}{401-1} + \dfrac{1}{241-1} \$ and \$ \dfrac{1}{101-1} = \dfrac{1}{601-1} + \dfrac{1}{571-1} + \dfrac{1}{457-1} + \dfrac{1}{229-1} \$

So now I can just replace the 19 with 71, 151, 101, 151, 211, 229, 241, 281, 401, 457, 471, 601 and the sequence will maintain it's properties.

I also discovered that I can replace 19 with the sequence 71, 151, 101, 151, 211, 241, 241, 281, 401, 433, 541, 601, but that has 241 twice.

After that improvement I also noticed that 79 could be replaced with 521, 313, 131, to increase the size by 2 more. And 73 can be replaced with 113, 379, 433 for another 2.

Score 32 34 36

{5, 7, 11, 13, 17, 23, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 101, 113, 131, 137, 151, 211, 229, 241, 281, 313, 379, 401, 433, 457, 491, 521, 571, 601, 25117, 293362609}

This is an improvement of Arnauld's answer. I just noticed that

\$ \dfrac{1}{19-1}=\dfrac{1}{73-1}+\dfrac{1}{61-1}+\dfrac{1}{41-1} \$

But 41 and 61 were already used in Arnauld's answer so I had to then figure out that

\$ \dfrac{1}{61-1} = \dfrac{1}{151-1} + \dfrac{1}{101-1} \$ and \$ \dfrac{1}{41-1} = \dfrac{1}{281-1}+\dfrac{1}{211-1}+\dfrac{1}{151-1}+\dfrac{1}{101-1} \$

But now I am using 151 and 101 twice. So I spent some time and discovered that

\$ \dfrac{1}{151-1} = \dfrac{1}{401-1} + \dfrac{1}{241-1} \$ and \$ \dfrac{1}{101-1} = \dfrac{1}{601-1} + \dfrac{1}{571-1} + \dfrac{1}{457-1} + \dfrac{1}{229-1} \$

So now I can just replace the 19 with 71, 151, 101, 151, 211, 229, 241, 281, 401, 457, 471, 601 and the sequence will maintain it's properties.

I also discovered that I can replace 19 with the sequence 71, 151, 101, 151, 211, 241, 241, 281, 401, 433, 541, 601, but that has 241 twice.

After that improvement I also noticed that 79 could be replaced with 521, 313, 131, to increase the size by 2 more. And 73 can be replaced with 113, 379, 433 for another 2.

Score 32 34

[5, 7, 11, 13, 17, 23, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 101, 131, 137, 151, 211, 229, 241, 281, 313, 401, 457, 491, 521, 571, 601, 25117, 293362609]

This is an improvement of Arnauld's answer. I just noticed that

\$ \dfrac{1}{19-1}=\dfrac{1}{73-1}+\dfrac{1}{61-1}+\dfrac{1}{41-1} \$

But 41 and 61 were already used in Arnauld's answer so I had to then figure out that

\$ \dfrac{1}{61-1} = \dfrac{1}{151-1} + \dfrac{1}{101-1} \$ and \$ \dfrac{1}{41-1} = \dfrac{1}{281-1}+\dfrac{1}{211-1}+\dfrac{1}{151-1}+\dfrac{1}{101-1} \$

But now I am using 151 and 101 twice. So I spent some time and discovered that

\$ \dfrac{1}{151-1} = \dfrac{1}{401-1} + \dfrac{1}{241-1} \$ and \$ \dfrac{1}{101-1} = \dfrac{1}{601-1} + \dfrac{1}{571-1} + \dfrac{1}{457-1} + \dfrac{1}{229-1} \$

So now I can just replace the 19 with 71, 151, 101, 151, 211, 229, 241, 281, 401, 457, 471, 601 and the sequence will maintain it's properties.

I also discovered that I can replace 19 with the sequence 71, 151, 101, 151, 211, 241, 241, 281, 401, 433, 541, 601, but that has 241 twice.

After that improvement I also noticed that 79 could be replaced with 521, 313, 131, to increase the size by 2 more.

Score 32 34 36

{5, 7, 11, 13, 17, 23, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 101, 113, 131, 137, 151, 211, 229, 241, 281, 313, 379, 401, 433, 457, 491, 521, 571, 601, 25117, 293362609}

This is an improvement of Arnauld's answer. I just noticed that

\$ \dfrac{1}{19-1}=\dfrac{1}{73-1}+\dfrac{1}{61-1}+\dfrac{1}{41-1} \$

But 41 and 61 were already used in Arnauld's answer so I had to then figure out that

\$ \dfrac{1}{61-1} = \dfrac{1}{151-1} + \dfrac{1}{101-1} \$ and \$ \dfrac{1}{41-1} = \dfrac{1}{281-1}+\dfrac{1}{211-1}+\dfrac{1}{151-1}+\dfrac{1}{101-1} \$

But now I am using 151 and 101 twice. So I spent some time and discovered that

\$ \dfrac{1}{151-1} = \dfrac{1}{401-1} + \dfrac{1}{241-1} \$ and \$ \dfrac{1}{101-1} = \dfrac{1}{601-1} + \dfrac{1}{571-1} + \dfrac{1}{457-1} + \dfrac{1}{229-1} \$

So now I can just replace the 19 with 71, 151, 101, 151, 211, 229, 241, 281, 401, 457, 471, 601 and the sequence will maintain it's properties.

I also discovered that I can replace 19 with the sequence 71, 151, 101, 151, 211, 241, 241, 281, 401, 433, 541, 601, but that has 241 twice.

After that improvement I also noticed that 79 could be replaced with 521, 313, 131, to increase the size by 2 more. And 73 can be replaced with 113, 379, 433 for another 2.

Score 32

[5, 7, 11, 13, 17, 23, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 101, 137, 151, 211, 229, 241, 281, 401, 457, 491, 571, 601, 25117, 293362609]

This is an improvement of Arnauld's answer. I just noticed that

\$ \dfrac{1}{19-1}=\dfrac{1}{73-1}+\dfrac{1}{61-1}+\dfrac{1}{41-1} \$

But 41 and 61 were already used in Arnauld's answer so I had to then figure out that

\$ \dfrac{1}{61-1} = \dfrac{1}{151-1} + \dfrac{1}{101-1} \$ and \$ \dfrac{1}{41-1} = \dfrac{1}{281-1}+\dfrac{1}{211-1}+\dfrac{1}{151-1}+\dfrac{1}{101-1} \$

But now I am using 151 and 101 twice. So I spent some time and discovered that

\$ \dfrac{1}{151-1} = \dfrac{1}{401-1} + \dfrac{1}{241-1} \$ and \$ \dfrac{1}{101-1} = \dfrac{1}{601-1} + \dfrac{1}{571-1} + \dfrac{1}{457-1} + \dfrac{1}{229-1} \$

So now I can just replace the 19 with 71, 151, 101, 151, 211, 229, 241, 281, 401, 457, 471, 601 and the sequence will maintain it's properties.

I also discovered that I can replace 19 with the sequence 71, 151, 101, 151, 211, 241, 241, 281, 401, 433, 541, 601, but that has 241 twice.

Score 32 34

[5, 7, 11, 13, 17, 23, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 101, 131, 137, 151, 211, 229, 241, 281, 313, 401, 457, 491, 521, 571, 601, 25117, 293362609]

This is an improvement of Arnauld's answer. I just noticed that

\$ \dfrac{1}{19-1}=\dfrac{1}{73-1}+\dfrac{1}{61-1}+\dfrac{1}{41-1} \$

But 41 and 61 were already used in Arnauld's answer so I had to then figure out that

\$ \dfrac{1}{61-1} = \dfrac{1}{151-1} + \dfrac{1}{101-1} \$ and \$ \dfrac{1}{41-1} = \dfrac{1}{281-1}+\dfrac{1}{211-1}+\dfrac{1}{151-1}+\dfrac{1}{101-1} \$

But now I am using 151 and 101 twice. So I spent some time and discovered that

\$ \dfrac{1}{151-1} = \dfrac{1}{401-1} + \dfrac{1}{241-1} \$ and \$ \dfrac{1}{101-1} = \dfrac{1}{601-1} + \dfrac{1}{571-1} + \dfrac{1}{457-1} + \dfrac{1}{229-1} \$

So now I can just replace the 19 with 71, 151, 101, 151, 211, 229, 241, 281, 401, 457, 471, 601 and the sequence will maintain it's properties.

I also discovered that I can replace 19 with the sequence 71, 151, 101, 151, 211, 241, 241, 281, 401, 433, 541, 601, but that has 241 twice.

After that improvement I also noticed that 79 could be replaced with 521, 313, 131, to increase the size by 2 more.