A306825 Primitive part of A001353(n).

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

Offset: 1

Author

Jianing Song, Mar 16 2019

Keywords

• nonn

Comments

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. [By the formula below, there is no such p. - Jianing Song, Oct 31 2024]

Examples

			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.
		

Links

• Eric Weisstein's World of Mathematics, Sylvester Cyclotomic Number
• Wikipedia, Lucas Lehmer Primality Test

Crossrefs

Cf. A001353, A238490, A309040.

Similar sequences: A061446, A008555.

Programs

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]))

Formula

Product_{d|n} a(d) = A001353(n), that is, a(n) = A001353(n)/(Product_{dA001353(d)^mu(n/d), where mu = A008683.

a(n) = A309040(2*n) for even n; A309040(n)*A309040(2*n) for odd n > 1. - Jianing Song, Oct 31 2024