a(2) = 12 because 5 (= A345977(2)) + 7 = 12 = 2^2*3 is the first sum S of 2 consecutive primes for which omega(S) = 2. 2 + 3 = 5, and 3 + 5 = 8 = 2^3 both have only one prime dividing S.
a(3) = 1015 = 331 + 337 + 347 = 5*7*29;
a(4) = 390 = 89 + 97 + 101 + 103 = 2*3*5*13;
a(5) = 26565 = 5297 + 5303 + 5309 + 5323 + 5333 = 3*5*7*11*23.
Since a(n) is the sum of n consecutive primes, a(n) is even iff n is even (provided that 2 is not among the consecutive primes, which happens only at n=1).
It seems that a(n) is usually squarefree, but a(2) and a(14) are multiples of 4.
The prime factorizations of the first 19 terms are shown in the table below. (To highlight the tendency of the terms to include the smallest odd primes among their divisors, each prime < 80 has its own column.)
.
n prime factorization of a(n) primes < 80 | > 80
-- ---------------------------------------------------------------+-----
1 2 |
2 2^2*3 |
3 5*7 *29 |
4 2*3*5 *13 |
5 3*5*7*11 *23 |
6 2*3*5*7*11 *29 |
7 3*5*7 *13*17*19 *47 |
8 2*3*5*7*11*13*17 |*97
9 3*5*7*11*13 *19*23 *37 *61 |
10 2*3*5*7*11*13*17*19 *29 *53 |
11 3*5*7*11*13*17 *23*29*31*37 *43 |
12 2*3*5*7*11*13*17*19*23*29 *43 *73 |
13 3*5*7*11*13*17*19*23*29*31 *43 *59 *73 |
14 2^2*3*5*7*11*13*17*19*23*29*31 *41 *53 *67 |
15 3*5*7*11*13*17*19*23*29*31 *41*43 *53*59 *73 |
16 2*3*5*7*11*13*17*19*23*29*31*37 *43 *53 *67*71 |
17 3*5*7*11*13*17*19*23*29*31*37*41*43 *53*59 *73 |*107
18 2*3*5*7*11*13*17*19*23 *31*37*41*43*47*53*59*61*67 |
19 3*5*7*11*13*17*19*23*29*31*37*41*43*47 *59*61 *71 *79|*131
(End)
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