A211204 a(1) = 2; for n > 1, a(n) > a(n-1) is the smallest prime for which the set {a(1), a(2), ..., a(n)} lacks at least one residue modulo every odd prime less than or equal to a(n).

2, 3, 5, 11, 17, 23, 41, 47, 53, 71, 83, 101, 107, 113, 131, 137, 167, 173, 191, 197, 233, 251, 257, 263, 311, 317, 347, 353, 401, 431, 443, 461, 467, 503, 521, 563, 593, 641, 647, 653, 677, 683, 701, 743, 761, 773, 797, 827, 857, 863, 881, 911, 941, 947, 971

Offset: 1

Author

Vladimir Shevelev and Peter J. C. Moses, Feb 04 2013

Keywords

• nonn

Comments

By construction, for every odd prime p > 1, the sequence does not contain a full residue system modulo p. For n >= 4, all differences a(n) - a(n-1) are multiples of 6; otherwise said, a(n) == 5 (mod 6).

Conjecture: The sequence contains infinitely many "twins" with a(n)-a(n-1) = 6.

All terms greater than 3 are 2 mod 3, so the sequence does not contain a complete residue system mod 3; all terms are not 4 mod 5, so the sequence does not contain a complete residue system mod 5; since 7 is absent in the sequence, there is not a complete residue system mod 7.

By the Chinese remainder theorem and Dirichlet's theorem on arithmetic progressions, the sequence is infinite. - Dimiter Skordev, Apr 05 2022

Links

• Dimiter Skordev, Table of n, a(n) for n = 1..1000 (first 92 terms from Vladimir Shevelev and Peter J. C. Moses)

Crossrefs

Cf. A210537.

Programs

Python

def isPrime(n):
    if (n%2==0): return n==2
    for i in range(3,int(n**0.5+1),2):
        if (n%i==0): return False
    return n>1
def nextPrime(n):
    n=n+1
    while not isPrime(n): n=n+1
    return n
def a(n):
    p,L,S=2,[],[]
    while len(L)Dimiter Skordev, Apr 05 2022

Extensions

Edited by M. F. Hasler, Feb 13 2013