login
A327653
Composite numbers k coprime to 13 such that k divides A006190(k-Kronecker(13,k)).
13
10, 119, 649, 1189, 1763, 3599, 4187, 5559, 6681, 12095, 12403, 12685, 12871, 12970, 14041, 14279, 15051, 16109, 19043, 22847, 23479, 24769, 26795, 28421, 30743, 30889, 31631, 31647, 33919, 34997, 37949, 38503, 39203, 41441, 46079, 48577, 49141, 50523, 50545, 53301, 56279, 58081, 58589
OFFSET
1,1
COMMENTS
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n) = m*x(n-1) + x(n-2) for k >= 2. For primes p, we have (a) p divides x(p-((m^2+4)/p); (b) x(p) == ((m^2+4)/p) (mod p), where (D/p) is the Kronecker symbol. This sequence gives composite numbers k such that gcd(k, m^2+4) = 1 and that a condition similar to (a) holds for k, where m = 3.
If k is not required to be coprime to m^2 + 4 (= 13), then there are 360 such k <= 10^5 and 1506 such k <= 10^6, while there are only 62 terms <= 10^5 and 197 terms <= 10^6 in this sequence.
Also composite numbers k coprime to 13 such that A322907(k) divides k - Kronecker(13,k).
LINKS
EXAMPLE
A006190(9) = 12970 which is divisible by 10, so 10 is a term.
PROG
(PARI) seqmod(n, m)=((Mod([3, 1; 1, 0], m))^n)[1, 2]
isA327653(n)=!isprime(n) && !seqmod(n-Kronecker(13, n), n) && gcd(n, 13)==1 && n>1
CROSSREFS
m m=1 m=2 m=3
k | x(k-Kronecker(m^2+4,k))* A081264 U A141137 A327651 this seq
k | x(k)-Kronecker(m^2+4,k) A049062 A099011 A327654
* k is composite and coprime to m^2 + 4.
Cf. A006190, A322907, A011583 ({Kronecker(13,n)}).
Sequence in context: A145510 A160601 A338975 * A024128 A289215 A223010
KEYWORD
nonn
AUTHOR
Jianing Song, Sep 20 2019
STATUS
approved