OFFSET
1,1
COMMENTS
Conjecture: for any prime number p, there exists at least one prime number q such that the sequence of p-1 consecutive prime numbers {p_k, p_{k+1},...,p_{k+p-2}} with p_k = q forms a complete system of nonzero residues modulo p.
REFERENCES
P. G. L. Dirichlet, Beweis des Satzes ... unendlich viele Primzahlen enthält, Abh. Königl. Preuß. Akad. Wiss., 1837, pp. 45-81.
G. H. Hardy, J. E. Littlewood, Some problems of "Partitio numerorum"; III: On the expression of a number as a sum of primes, Acta Math. 44 (1923), pp. 1-70.
J. Maynard, Small gaps between primes, Ann. of Math. 181 (2015), pp. 383-413.
EXAMPLE
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p q [p-1 consecutive primes] {residues modulo p}
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2 3 [3] {1}
3 5 [5, 7] {1, 2}
5 11 [11, 13, 17, 19] {1, 2, 3, 4}
7 11 [11, 13, 17, 19, 23, 29] {1,..., 6}
11 41 [41, 43,..., 79] {1,..., 10}
13 10267 [10267,..., 10343] {1,..., 12}
MAPLE
restart;
findPQ := proc(limit_p)
local p, q, residues, current_q, i, target;
p := 2;
while p <= limit_p do
target := {seq(i, i=1..p-1)};
q := 2;
while true do
residues := {};
current_q := q;
for i from 1 to p-1 do
residues := residues union {current_q mod p};
current_q := nextprime(current_q);
end do;
if residues = target then
printf("p = %d, q = %d\n", p, q);
break;
end if;
q := nextprime(q);
end do;
p := nextprime(p);
end do;
end proc:
findPQ(17);
PROG
(PARI) isok(p, q) = my(list=List); listput(list, q); for (i=2, p-1, q=nextprime(q+1); listput(list, q)); my(v=vector(p-1, k, list[k] % p)); vecmin(v) && (#Set(v) == p-1);
a(n) = my(p=prime(n), q=nextprime(p+1)); while (!isok(p, q), q=nextprime(q+1)); q; \\ Michel Marcus, Mar 03 2026
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Michel Lagneau, Mar 03 2026
EXTENSIONS
a(7)-a(9) corrected by Michel Marcus, Mar 06 2026
a(10)-a(11) from Michael S. Branicky, Mar 22 2026
STATUS
approved