OFFSET
1,1
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).
LINKS
Pontus von Brömssen, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals)
EXAMPLE
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.
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Pontus von Brömssen, Jan 18 2025
STATUS
approved
