Question 1209831
Let S={1,2,3,…,12}.
The primes in S are P={2,3,5,7,11}.
We want to find the number of subsets of S such that no two consecutive primes are members.

Let's consider the primes in S. We want to choose a subset of P such that no two consecutive primes are chosen. Let a 
n
​
  be the number of subsets of {p 
1
​
 ,p 
2
​
 ,…,p 
n
​
 } with no consecutive primes, where p 
i
​
  is the i-th prime.

For n=0, a 
0
​
 =1 (empty set).
For n=1, a 
1
​
 =2 (empty set, {p 
1
​
 }).
For n=2, a 
2
​
 =3 (empty set, {p 
1
​
 }, {p 
2
​
 }).
For n=3, a 
3
​
 =5 (empty set, {p 
1
​
 }, {p 
2
​
 }, {p 
3
​
 }, {p 
1
​
 ,p 
3
​
 }).
For n=4, a 
4
​
 =8 (empty set, {p 
1
​
 }, {p 
2
​
 }, {p 
3
​
 }, {p 
4
​
 }, {p 
1
​
 ,p 
3
​
 }, {p 
1
​
 ,p 
4
​
 }, {p 
2
​
 ,p 
4
​
 }).
For n=5, a 
5
​
 =13 (empty set, {p 
1
​
 }, {p 
2
​
 }, {p 
3
​
 }, {p 
4
​
 }, {p 
5
​
 }, {p 
1
​
 ,p 
3
​
 }, {p 
1
​
 ,p 
4
​
 }, {p 
1
​
 ,p 
5
​
 }, {p 
2
​
 ,p 
4
​
 }, {p 
2
​
 ,p 
5
​
 }, {p 
3
​
 ,p 
5
​
 }, {p 
1
​
 ,p 
3
​
 ,p 
5
​
 }).

We observe that a 
n
​
 =F 
n+2
​
 , where F 
n
​
  is the n-th Fibonacci number (F 
0
​
 =0,F 
1
​
 =1,F 
2
​
 =1,F 
3
​
 =2,F 
4
​
 =3,F 
5
​
 =5,…).

Thus, the number of subsets of P with no consecutive primes is a 
5
​
 =F 
7
​
 =13.

Now consider the non-prime numbers in S: N={1,4,6,8,9,10,12}.
The number of subsets of N is 2 
7
 =128.

Each subset of N can be combined with any subset of P that has no consecutive primes.
Thus, the total number of subsets of S with no consecutive primes is 13×128=1664.

Final Answer: The final answer is  
1664
​