For this problem you can use three array primeNum[](contain all prime number between 1 to 32000), diff[] (contain difference between current prime number and previous prime number), and element[] (contain the number of element at same difference).

1. First pre-generate all prime number from 1 to 32000 and store to array primeNum[].

2. Set difference = current prime number – previous prime number . (i.e. diff[i] = primeNum[i]-primeNum[i-1]).

3. Set element

      A. If (diff[i-1]!=diff[i]) then set element[i]=2

      B. If (diff[i-1]==diff[i]) then set element[i] = element[i-1]+1

4.   For input x and y

Find the start index for x. If primeNum[i]>=x then set start_index = i

For  i=star_index ;  primeNum[i]<=y ; i++

                If element[i]>=3 and element[i+1]!=element[i] and primeNum[i-element[i]+1]>=x

                Then print all prime from primeNum[i-element[i]+1] to primeNum[i].

Remember,  y can be greater than x. If y is greater than x then swap(x,y)

Example:

Prime number between 1 and 20 are

primeNum[1]= 2       diff[1]=2      element[1]=2

primeNum[2]= 3       diff[2]=1      element[2]=2

primeNum[3]= 5       diff[3]=2      element[3]=2

primeNum[4]= 7       diff[4]=2      element[4]=3

primeNum[5]=11       diff[5]=4      element[5]=2

priemNum[6]=13       diff[6]=2      element[6]=2

primeNum[7]=17       diff[7]=4      element[7]=2

primeNum[8]=19       diff[8]=2      element[8]=2

And so on.

a. If x=1 and y=18, then start_index=1  and end_index=7

     Form i=1 to 7 , for i=4 element[4]>=3 so find the lower boundary which 3 (primeNum[4-3+1]) , and 3 is     greater or equal x=2. So, the sequence 3 5 7

     b. If x=18 and y=4, then swap(x,y) after that x=4 and y=18. For that start_index=3 and end_index=7

     Form i=3 to 7 , for i=4 element[4]>=3 so find the lower boundary which 3 (primeNum[4-3+1]) , and 3 is not greater or equal  x=4. So, do not print the sequence.