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A327654
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Composite numbers k coprime to 13 such that k divides A006190(k) - Kronecker(13,k).
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4
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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A006190(8) = 3927 == Kronecker(13,8) (mod 8), so 8 is a term.
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PROG
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(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
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CROSSREFS
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m m=1 m=2 m=3
* k is composite and coprime to m^2 + 4.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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