A380277 A version of the array A229607 without duplicates, read by antidiagonals: each row starts with the least prime not in a previous row, and each number p in a row is followed by the greatest prime q in the interval p < q < 2*p not in a previous row (or 0 if no such q exists).

2, 3, 11, 5, 19, 17, 7, 37, 31, 29, 13, 73, 61, 53, 41, 23, 139, 113, 103, 79, 47, 43, 277, 223, 199, 157, 89, 59, 83, 547, 443, 397, 313, 173, 109, 67, 163, 1093, 883, 787, 619, 337, 211, 131, 71, 317, 2179, 1759, 1571, 1237, 673, 421, 257, 137, 97

Offset: 1

Author

Pontus von Brömssen, Jan 18 2025

Keywords

• nonn
• tabl

Comments

It appears that the first column is A104272.

Proof: (This proof assumes that all terms of the array are nonzero. It would be nice to see a proof of this.) Let n >= 2 and p = T(n,1). To prove that p = A104272(n) we need to prove that pi(x)-pi(x/2) >= n for x >= p and that pi(p-1)-pi((p-1)/2) < n. Let x >= p and let q be the smallest prime larger than x. In each of the rows 1..n there are consecutive terms r < r' with r < q <= r' < 2*r, so q/2 < r < q. Hence there are at least n primes between q/2 and q (not counting q itself), i.e., pi(q)-pi(q/2) >= n+1. It follows that pi(x)-pi(x/2) = pi(q)-1-pi(x/2) >= pi(q)-1-pi(q/2) >= n. Finally, if pi(p-1)-pi((p-1)/2) >= n there would exist two consecutive terms r and r' in one of the rows 1..(n-1) with (p-1)/2 < r < r' <= p-1. This is impossible, because then p (or some larger prime) would have been chosen instead of r' as the successor of r. Hence pi(p-1)-pi((p-1)/2) < n. This concludes the proof (with the caveat above).

Examples

			Array starts:
   2,   3,   5,   7,   13,   23,   43,    83,   163,   317, ...
  11,  19,  37,  73,  139,  277,  547,  1093,  2179,  4357, ...
  17,  31,  61, 113,  223,  443,  883,  1759,  3517,  7027, ...
  29,  53, 103, 199,  397,  787, 1571,  3137,  6271, 12541, ...
  41,  79, 157, 313,  619, 1237, 2473,  4943,  9883, 19763, ...
  47,  89, 173, 337,  673, 1327, 2647,  5281, 10559, 21107, ...
  59, 109, 211, 421,  839, 1669, 3331,  6661, 13313, 26597, ...
  67, 131, 257, 509, 1013, 2017, 4027,  8053, 16103, 32203, ...
  71, 137, 271, 541, 1069, 2137, 4273,  8543, 17077, 34147, ...
  97, 193, 383, 761, 1511, 3019, 6037, 12073, 24137, 48271, ...
  ...
The least prime not in any of the first 6 rows is T(7,1) = 59. The greatest prime less than 2*59 = 118 is 113, but that number appears in a previous row as T(3,4). The next smaller prime is 109, which does not appear in a previous row, so T(7,2) = 109.
		

Links

• Pontus von Brömssen, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals)

Crossrefs

Cf. A000720, A006992 (first row), A104272, A229607.