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A297620 Positive numbers n such that n^2 == p (mod q) and n^2 == q (mod p) for some consecutive primes p,q. 1
6, 10, 12, 24, 42, 48, 62, 72, 76, 84, 90, 93, 108, 110, 120, 122, 128, 145, 146, 174, 187, 188, 194, 204, 208, 215, 220, 228, 232, 240, 241, 264, 297, 306, 310, 314, 317, 326, 329, 336, 349, 357, 366, 372, 386, 408, 410, 423, 426, 431, 444, 454, 456, 468, 470, 474, 518, 522, 535, 538, 546, 548 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Positive numbers n such that n^2 == p+q mod (p*q) for some consecutive primes p, q.
Each pair of consecutive primes p,q such that p is a quadratic residue mod q and p and q are not both == 3 (mod 4) contributes infinitely many members to the sequence.
Odd terms of this sequence are 93, 145, 187, 215, 241, 297, 317, 329, 349, 357, 423, 431, 535, ... - Altug Alkan, Jan 01 2018
LINKS
EXAMPLE
a(3) = 12 is in the sequence because 71 and 73 are consecutive primes with 12^2 == 73 (mod 71) and 12^2 == 71 (mod 73).
MAPLE
N:= 1000: # to get all terms <= N
R:= {}:
q:= 3:
while q < N^2 do
p:= q;
q:= nextprime(q);
if ((p mod 4 <> 3) or (q mod 4 <> 3)) and numtheory:-quadres(q, p) = 1 then
xp:= numtheory:-msqrt(q, p); xq:= numtheory:-msqrt(p, q);
for sp in [-1, 1] do for sq in [-1, 1] do
v:= chrem([sp*xp, sq*xq], [p, q]);
R:= R union {seq(v+k*p*q, k = 0..(N-v)/(p*q))}
od od;
fi;
od:
sort(convert(R, list));
CROSSREFS
Contains A074924.
Sequence in context: A046363 A101086 A342648 * A074924 A064166 A373678
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified August 8 18:48 EDT 2024. Contains 375023 sequences. (Running on oeis4.)