A327654 Composite numbers k coprime to 13 such that k divides A006190(k) - Kronecker(13,k).

4, 8, 9, 119, 399, 649, 1023, 1179, 1189, 1199, 1881, 2703, 3519, 4081, 4187, 5151, 7055, 7361, 10349, 12871, 13833, 14041, 15519, 16109, 18639, 22593, 23479, 24769, 26937, 28421, 29007, 31631, 34111, 34997, 38503, 41441, 44671, 48577, 50545, 51711, 53823, 56279, 57407, 58081, 59081

Offset: 1

Author

Jianing Song, Sep 20 2019

Keywords

• nonn

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 (b) holds for k, where m = 3.

If k is not required to be coprime to m^2 + 4 (= 13), then there are 352 such k <= 10^5, and 1457 such k <= 10^6, while there are only 54 terms <= 10^5 and 148 terms <= 10^6 in this sequence.

Examples

			A006190(8) = 3927 == Kronecker(13,8) (mod 8), so 8 is a term.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..2000

Crossrefs

m m=1 m=2 m=3

k | x(k-Kronecker(m^2+4,k))* A081264 U A141137 A327651 A327653

k | x(k)-Kronecker(m^2+4,k) A049062 A099011 this seq

both A212424 A327652 A327655

* k is composite and coprime to m^2 + 4.

Cf. A006190, A011583 ({Kronecker(13,n)}).

Programs

PARI

seqmod(n, m)=((Mod([3, 1; 1, 0], m))^n)[1, 2]
isA327654(n)=!isprime(n) && seqmod(n, n)==kronecker(13,n) && gcd(n,13)==1 && n>1