A360228 a(n) is the least prime p such that the primes from prime(n) to p contain a complete set of residues modulo at least one of these primes.

3, 7, 19, 29, 71, 103, 103, 191, 233, 317, 439, 439, 467, 467, 659, 659, 709, 1013, 1013, 1319, 1319, 1319, 1499, 1499, 1499, 1973, 1973, 2203, 2203, 2203, 3089, 3089, 3449, 3449, 3449, 3539, 3539, 3923, 3923, 4349, 4349, 4349, 4349, 4349, 4793, 4793, 4793, 4793, 5813, 5813, 5813, 5813, 5813

Offset

1,1

Links

Michael S. Branicky, Table of n, a(n) for n = 1..531 (terms 1..100 from Robert Israel)

Example

a(3) = 19 because prime(3) = 5 and the primes (5, 7, 11, 13, 17, 19} from 17 to 19 contain a complete set of residue mod 5:  5 == 0, 11 == 1, 7 == 2, 13 == 3 and 19 == 4 (mod 5).

Maple

P:= select(isprime, [2, seq(i, i=3..10^6, 2)]): nP:= nops(P):
f:= proc(i) local j, k, Q;
  Q:=Array(1..1,  j -> {$1..P[i+j-1]-1});
  for j from i+1 do
    for k from 1 to j-i do
      Q(k):= Q(k) minus {P[j] mod P[i+k-1]};
      if Q(k) = {} then return P[j] fi;
    od;
    Q(j-i+1):= {$1..P[j]-1} minus convert(P[i..j] mod P[j], set);
    if Q(j-i+1) = {} then return P[j] fi;
  od
end proc:
map(f, [$1..60]);

Prog

(Python)
from sympy import sieve
def a(n):
    pn = p = sieve[n]; res = dict(); i = 0
    while True:
        if p not in res:
            res[p] = {0} | {q%p for q in res}
        for q in res:
            res[q].add(p%q)
            if len(res[q]) == q:
                return p
        i += 1
        p = sieve[n+i]
print([a(n) for n in range(1, 54)]) # Michael S. Branicky, Feb 02 2023

Crossrefs

Cf. A358238.

Sequence in context: A056725 A091738 A340752 * A358238 A136054 A006032

Adjacent sequences: A360225 A360226 A360227 * A360229 A360230 A360231

Keyword

nonn

Author

Robert Israel, Jan 30 2023

Status

approved