Euler's expression f(n) = n^2  n + 41 generates 40 consecutive terms that are prime, but this sequence contains 44 consecutive terms that are prime.
The first differences of Euler's polynomial are the even numbers 2 and up. This sequence extends backwards for four further terms in a similar way:
terms: 13 23 31 37 41 43 47 53 61 71 ...
1st differences: 10 8 6 4 2 4 6 8 10 ...
2nd differences: 2 2 2 2 2 2 2 2 ...
So this sequence consists of two quadratic pieces: one with 2nd differences 2, the other with second differences +2.
The terms that are primes from Euler's expression are 41, 43, 47, 53, 61, ...; each of these can be obtained by adding to 41 the product of 2 and the sum of the first k nonnegative integers:
41 + 2*(0) = 41 (prime)
41 + 2*(0 + 1) = 43 (prime)
41 + 2*(0 + 1 + 2) = 47 (prime)
41 + 2*(0 + 1 + 2 + 3) = 53 (prime)
41 + 2*(0 + 1 + 2 + 3 + 4) = 61 (prime)
etc. The sum of the first k nonnegative integers is (k1)*k/2, and adding 41 to twice that sum gives 2*(k1)*k/2 + 41 = k^2  k + 41.
Also, for k = 1..5, subtracting 2 times the sum of the first k positive integers from 43 gives 5 primes in a row:
43  2*(1) = 41 (prime)
43  2*(1 + 2) = 37 (prime)
43  2*(1 + 2 + 3) = 31 (prime)
43  2*(1 + 2 + 3 + 4) = 23 (prime)
43  2*(1 + 2 + 3 + 4 + 5) = 13 (prime)
(The next term in the above sequence, 43  2*(1 + 2 + 3 + 4 + 5 + 6) = 1, is not a prime number.)
So this sequence is piecewise quadratic and consists of two overlapping quadratic subsequences: one contains the 5 terms 13, 23, 31, 37, 41, while the other contains the 40 terms from Euler's expression, i.e., 41, 43, 47, ..., 1601, and there is one overlapping term (41), for a total of 5 + 40  1 = 44 terms, all of which are prime.
