According to the meaning of the question, each line becomes arithmetic progression.

If the tolerances are d 1, d2, d3, …, d32, …, dn respectively, the tolerance is geometric series {dn}, and the general formula is dn = 2 n (2 to the nth power).

D32 = 2 32, so only the first number of each row is calculated. Let the first number of the nth row be an 1 (n > 1).

So a 21= a11+(a11+21).

a3 1=a2 1+(a2 1+2^2)

……

a(n- 1) 1=a(n-2) 1+[a(n-2) 1+2^(n-2)]

an 1=a(n- 1) 1+[a(n- 1) 1+2^(n- 1)]

Obviously, the sequence {an 1} has the recursive relation an1= a (n-1)1+[a (n-1)1+2 (n-/kloc-).

Yi dean1-n * 2 (n-1) = 2 [a (n-1)1+(n-1) * 2 (n-2)]

We find that an 1-n * 2 (n- 1) = 0, so an 1 = n * 2 (n- 1).

Then it is obtained from arithmetic progression's general formula.

The number k in the nth row is n * [2 (n-1)]+(2n) * (k-1) = 2 (n-1) * (n+2k-2).

So the number of line 32 17 is 64 * (2 3 1) = 2 37.