human
Solve in Python:
A sequence of number is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.

For example, these are arithmetic sequence:
1, 3, 5, 7, 9
7, 7, 7, 7
3, -1, -5, -9

The following sequence is not arithmetic. 1, 1, 2, 5, 7 

A zero-indexed array A consisting of N numbers is given. A slice of that array is any pair of integers (P, Q) such that 0 

A slice (P, Q) of array A is called arithmetic if the sequence:
    A[P], A[p + 1], ..., A[Q - 1], A[Q] is arithmetic. In particular, this means that P + 1 < Q.

The function should return the number of arithmetic slices in the array A. 

Example:

A = [1, 2, 3, 4]

return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] itself.
 ... Can you provide a the Python program you requested?
assistant
#Here is the Python program you requested.

class Solution:
     def numberOfArithmeticSlices(self, A):
         curr, sum = 0, 0
         for i in range(2,len(A)):
             if A[i]-A[i-1] == A[i-1]-A[i-2]:
                 curr += 1
                 sum += curr
             else:
                 curr = 0
         return sum
 #         solution = 0
 #         connected = 1
 #         old_diff = None
 #         sequences = []
 #         if len(A) < 3:
 #             return 0
         
 #         for index,num in enumerate(A):
 #             if index < len(A) - 1:
 #                 new_diff = num - A[index + 1]
 #             else:
 #                 new_diff = A[index - 1] - num
 #             if old_diff == new_diff:
 #                 if index == len(A) - 1 and connected >= 3:
 #                     connected += 1
 #                     sequences.append(connected)
 #                 connected += 1
 #             else:
 #                 old_diff = new_diff
 #                 if connected > 2:
 #                     sequences.append(connected)
 #                 connected = 1
 #         for sequence in sequences:
 #             prev = 0
 #             while sequence >= 2:
 #                 prev += 1
 #                 solution += prev
 #                 sequence -= 1
 #         return solution