A113460 Triangle read by rows: n-th row is the lexicographically earliest arithmetic progression of n numbers all having the same prime signature.
1, 2, 3, 3, 5, 7, 5, 11, 17, 23, 5, 11, 17, 23, 29, 7, 37, 67, 97, 127, 157, 7, 157, 307, 457, 607, 757, 907, 11, 1210241, 2420471, 3630701, 4840931, 6051161, 7261391, 8471621, 11, 32671181, 65342351, 98013521, 130684691, 163355861, 196027031, 228698201, 261369371
Offset: 1
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)
Extensions
Erroneous commas in sequence deleted by N. J. A. Sloane, Sep 22 2007
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