OFFSET
1,2
COMMENTS
Start of 4 consecutive terms in A057128. Note that there are no 5 consecutive terms there.
Two interesting properties:
(a) All terms are congruent to 1 mod 360.
Proof: since -3 is not a quadratic residue modulo any prime == 5 (mod 6) we have k == 1 (mod 5) and k == 0 or 1 (mod 6). If k is even, then k == 6 (mod 30). -3 is not a quadratic residue modulo 9 so k == 3 (mod 9), then k == 66 (mod 90), k + 3 = 90*t + 69 = 3*(30*t + 23) but 30*t + 23 == 5 (mod 6), so -3 is not a quadratic residue modulo 30*t + 23, a contradiction. Thus k must be odd, then k == 1 (mod 30). For the same reason k == 1 or 4 (mod 9). If k == 4 (mod 9), then k == 31 (mod 90), k + 2 = 90*t + 33 = 3*(30t + 11) but 30*t + 11 == 5 (mod 6), a contradiction. So k == 1 (mod 9), then k == 1 (mod 90). If k == 91 (mod 180), then k + 3 = 180*t + 94 = 2*(90*t + 47) but 90*t + 47 == 5 (mod 6), a contradiction. So k == 1 (mod 180). -3 is not a quadratic residue modulo 8 so k == 1, 2, 3 or 4 (mod 8), thus k == 1 (mod 360) which is what we wanted.
(b) k is a term iff -3 is a quadratic residue modulo k*(k + 1)*(k + 2)*(k + 3)/2.
Proof: "<=" is obvious, since k*(k + 1)*(k + 2)*(k + 3)/2 is multiple of k, k + 1, k + 2 and k + 3. "=>": Note that -3 is a quadratic residue modulo lcm(k, k + 1, k + 2, k + 3). Now we show that lcm(k, k + 1, k + 2, k + 3) = k*(k + 1)*(k + 2)*(k + 3)/2. If not, then k is a multiple of 3, but by (a) we have k == 1 (mod 3), a contradiction.
LINKS
Jianing Song, Table of n, a(n) for n = 1..10000 (using data from A318911)
FORMULA
a(n) = 360*A318911(n) + 1.
EXAMPLE
721 is a term since 93^2 == -3 (mod 721), 137^2 == -3 (mod 722), 210^2 == -3 (mod 723) and 97^2 == -3 (mod 724).
PROG
(PARI) isA057128(n) = issquare(Mod(-3, n));
isA318527(n) = isA057128(n)&&isA057128(n+1)&&isA057128(n+2)&&isA057128(n+3);
for(n=1, 100000, if(isA318527(n), print1(n, ", ")))
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Aug 30 2018
STATUS
approved