

A307434


a(n) is the smallest k such that the prank of (Z/kZ)* is 2, where p = prime(n) and (Z/kZ)* is the multiplicative group of integers modulo n.


0



8, 63, 275, 1247, 1541, 4187, 14111, 43739, 6533, 13747, 116003, 33227, 61337, 74563, 186497, 79501, 586343, 269011, 432821, 485357, 128627, 451091, 83333, 191351, 377719, 491063, 638189, 551051, 2617309, 359341, 1164083, 206981, 902831, 929633, 2134277
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OFFSET

1,1


COMMENTS

The prank of a finite abelian group G is equal to log_p(#{x belongs to G : x^p = 1}) where p is a prime number. By definition, a(n) is the smallest k such that x^p == 1 (mod k) has exactly p^2 solutions, p = prime(n).
a(n) exists for all n: by Dirichlet's theorem on arithmetic progressions, there exists a primes q congruent to 1 modulo p, in which case C_p X C_p is a subgroup of (Z/(q*p^2)Z)*, p = prime(n).
a(n) is the smallest k such that C_p X C_p is a subgroup of (Z/kZ)*, p = prime(n).
Also, a(n) is the smallest k such that there exists some x, y such that ord(x,k) = ord(y,k) = prime(n) and the set of powers of x and the set of powers of y modulo k have trivial intersection {1}, where ord(x,k) is the multiplicative order of x modulo k.
Let p, q be odd primes, then (Z/(q^e)Z)* has prank 1 if q == 1 (mod p), or q = p and e >= 2; 0 otherwise. As a result, for n >= 2, a(n) is of the form (q_1)*(q_2) or (q_1)*p^2 where q_1 and q_2 are the smallest two primes congruent to 1 modulo prime(n).
It seems that for n >= 4, a(n) is not divisible by prime(n)^2, that is, there exists at least two primes that are smaller than prime(n)^2 and congruent to 1 modulo prime(n).


LINKS

Table of n, a(n) for n=1..35.
Wikipedia, Multiplicative group of integers modulo n


FORMULA

a(1) = 8; a(n) = A307436(prime(n)) for n >= 2.


EXAMPLE

(Z/8Z)* = C_2 X C_2, in which the solutions to x^2 == 1 (mod 8) are x == 1, 3, 5, 7 (mod 8) (4 solutions);
(Z/63Z)* = C_6 X C_6, in which the solutions to x^3 == 1 (mod 63) are x == 1, 4, 16, 22, 25, 37, 43, 46, 58 (mod 63) (9 solutions);
(Z/275Z)* = C_10 X C_20, in which the solutions to x^5 == 1 (mod 275) are x == 1, 16, 26, 31, 36, 56, 71, 81, 86, 91, 111, 126, 136, 141, 146, 166, 181, 191, 196, 201, 221, 236, 246, 251, 256 (mod 275) (25 solutions).


PROG

(PARI) a(n) = if(n==1, 8, my(p=prime(n), i=0, q=0); for(k=1, +oo, if(isprime(2*k*p+1), i++; if(i==1, q=2*k*p+1)); if(i==2, return(q*min(p^2, 2*k*p+1)))))


CROSSREFS

Cf. A307436, A035095.
Sequence in context: A132506 A080270 A178392 * A282311 A085433 A171313
Adjacent sequences: A307431 A307432 A307433 * A307435 A307436 A307437


KEYWORD

nonn


AUTHOR

Jianing Song, Apr 08 2019


STATUS

approved



