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A306825
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Primitive part of A001353(n).
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2
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1, 4, 15, 14, 209, 13, 2911, 194, 2703, 181, 564719, 193, 7865521, 2521, 34945, 37634, 1525870529, 2701, 21252634831, 37441, 6779137, 489061, 4122901604639, 37633, 274758906449, 6811741, 19726764303, 7263361, 11140078609864049, 40321, 155161278879431551
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OFFSET
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1,2
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COMMENTS
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A prime p is called a unique-period prime in base b if there is no other prime q such that the period length of 1/q is equal to that of 1/p. If q = a(2p) = A001353(2*p)/(4*A001353(p)) = ((2 + sqrt(3))^p + (2 - sqrt(3))^p)/4 is prime (this happens for p = 3, 5, 7, 11, 13, 17, 19, 79, 151, 199, 233, 251, 317, ...), where p is an odd prime, then q is a unique-period prime in base b = (sqrt(12*q^2 - 3) - 1)/2 (1/q has period length 3) as well as in base b' = (sqrt(12*q^2 - 3) + 1)/2 (1/q has period length 6). For example, a(6) = 13 is prime, so 13 is the only prime whose reciprocal has period length 3 in base 22 and the only prime whose reciprocal has period length 6 in base 23. Compare: If q = A000129(p) = A008555(p), then q is a unique-period prime in base b = sqrt(2*q^2 - 1) (1/q has period length 4).
By Lucas-Lehmer test, p is a Mersenne prime > 3 if and only if the smallest k such that p divides a(k) is k = (p - 1)/2.
For primes p, p^2 divides a(k) for some k if and only if p = 2 or p is in A238490. If p > 2, the only possible values for k are the divisors of (p - Legendre(3,p))/2 (e.g., 103^2 divides a(52) = 53028360515521 = 103^2 * 4998431569).
Conjecturally there must be infinitely many primes p such that a(p) is prime, but no such p is known.
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LINKS
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FORMULA
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Product_{d|n} a(d) = A001353(n), that is, a(n) = A001353(n)/(Product_{d<n, d|n} a(d)). Equivalently, a(n) = Product_{d|n} A001353(d)^mu(n/d), where mu = A008683.
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EXAMPLE
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For n = 8 we have: a(1) = A001353(1), a(1)*a(2) = A001353(2), a(1)*a(2)*a(4) = A001353(4), a(1)*a(2)*a(4)*a(8) = A001353(8). The solution is a(1) = 1, a(2) = 4, a(4) = 14 and a(8) = 194.
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PROG
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(PARI) b(n) = if(n==1, [1], my(v=vector(n)); v[1]=1; v[2]=4; for(i=3, n, v[i]=4*v[i-1]-v[i-2]); v)
a(n) = my(d=divisors(n)); prod(i=1, #d, (b(n)[d[i]])^moebius(n/d[i]))
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CROSSREFS
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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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