cp's OEIS Frontend

Examples

			Triangle begins:
  1;
  2,   3;
  3,   5,   7;
  5,  11,  17,  23;
  5,  11,  17,  23,  29;
  7,  37,  67,  97, 127, 157;
  ...
From _M. F. Hasler_, Oct 10 2024: (Start)
For row 1, we can take 1, which is the only integer to have prime signature {}.
For row 2, we can't use 1 (no two integers with that prime signature), but primes 2 & 3 are a valid and then also minimal choice.
For row 3, primes {3, 5, 7} are a valid choice and also smallest: we can't use 1, nor 2, for reasons of parity: the next prime would be odd but the third term of the arithmetic progression would then again be even and not prime.
  The same reasoning also excludes any higher power 2^m as starting term, which would require the same (m-th) power of odd primes as subsequent terms.
For rows 4 and 5, we can't start with the prime 3, because the 4th term of any arithmetic progression starting with 3 is again divisible by 3. Also 4 = 2^2 is excluded, see above. Thus, 5 is the smallest possible starting term for n = 4 and 5.
For row 6 and 7, we again can't start with a prime < nextprime(6) = 7, because there can't be more than 5 primes in AP starting with 5: the sixth term would always be divisible by 5 again. To start with the even semiprime 6 = 2*3 would require an AP of even semiprimes. Dividing by 2, we would have an AP of 6 primes starting with 3, which is impossible.(*) So, 7 is the smallest possibility.
(* This actually excludes all even semiprimes 2*p between prime(k-1) and prime(k) from being a starting term of a row in that range, because that would yield an AP of >= prime(k-1) primes starting with p < prime(k)/2 < prime(k-1), impossible.)
Rows 8 through 11 can't start with a prime < nextprime(8) = 11, as before. We have also excluded any 2^m and 2*3 as starting value. Starting with 9 = 3^2 would require an AP of squares of primes, but all larger squares of primes have a difference (6k +- 1)^2 - (6m +- 1)^2 divisible by 12, which is not the case for the difference with 3^2 = 9. The even semiprime 10 = 2*5 was also excluded above (*). Therefore, 11 is the smallest possible initial term. And so on. (End)