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 A098058 Prime(n) such that 4 does not divide the difference between prime(n) and prime(n+1). 17
 2, 3, 5, 11, 17, 23, 29, 31, 41, 47, 53, 59, 61, 71, 73, 83, 101, 107, 113, 131, 137, 139, 149, 151, 157, 167, 173, 179, 181, 191, 197, 227, 233, 239, 241, 251, 257, 263, 269, 271, 281, 283, 293, 311, 317, 331, 337, 347, 353, 367, 373, 383, 409, 419, 421, 431 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS First differences are also not divisible by 4. - Zak Seidov, Jun 23 2015 Starting with 3, group the primes into runs of consecutive primes either all == 1 (mod 4) or all == 3 (mod 4). Only the last prime of each run appears in this sequence. Since the runs alternate == 1 (mod 4) and == 3 (mod 4), so do the members of this sequence. - Franklin T. Adams-Watters, Jun 23 2015 The sequence is infinite, by Dirichlet's theorem on primes in arithmetic progressions. The sequence contains arbitrarily long gaps, by Daniel Shiu's theorem on strings of congruent primes (see A057619 and A057622). Conjecture: The sequence contains arbitrarily long strings of consecutive primes (see A289118). - Jonathan Sondow, Jun 25 2017 REFERENCES R. K. Guy, Unsolved Problems in Number Theory, A4. LINKS Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 Jens Kruse Andersen, Consecutive Congruent Primes EXAMPLE Prime(2) = 3, prime(3) = 5. 4 does not divide 5-3 so prime(2)=3 is in the sequence. Runs: (3), (5), (7,11), (17), (19, 23), (29), (31), (37,41), (43,47), (53), ... The sequence is 2 followed by the last member of each run. Differences within each run are always divisible by 4. MATHEMATICA Prime[Select[Range[100], Mod[Prime[ # + 1] - Prime[ # ], 4] !=0 &]] (* Ray Chandler, Oct 09 2006 *) PROG (PARI) f(n) = for(x=1, n, z=(prime(x+1)-prime(x)); if(z%4, print1(prime(x)", "))) (PARI) alist(n)=my(r=vector(n), p=2, np, k=0); while(k

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Last modified June 23 23:00 EDT 2021. Contains 345402 sequences. (Running on oeis4.)