Hello, I was just looking for feedback, comments, observations, on this potential new series:
Thanks,
Martin
Prime numbers that can be written as half the sum of three squares in exactly one unique way.
2, 3, 5, 7, 11, 23, 29, 71
There are only eight terms and no prime p > 71 has a unique solution due to class number growth restrictions under the Stark-Heegner theorem.
{ prime p : 2*p in { A094942 } }.
2: (2^2 + 0^2 + 0^2) / 2 = (4 + 0 + 0) / 2 = 2
3: (2^2 + 1^2 + 1^2) / 2 = (4 + 1 + 1) / 2 = 3
5: (3^2 + 1^2 + 0^2) / 2 = (9 + 1 + 0) / 2 = 5
7: (3^2 + 2^2 + 1^2) / 2 = (9 + 4 + 1) / 2 = 7
11: (3^2 + 3^2 + 2^2) / 2 = (9 + 9 + 4) / 2 = 11
23: (6^2 + 3^2 + 1^2) / 2 = (36 + 9 + 1) / 2 = 23
29: (7^2 + 3^2 + 0^2) / 2 = (49 + 9 + 0) / 2 = 29
71: (9^2 + 6^2 + 5^2) / 2 = (81 + 36 + 25) / 2 = 71