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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A307434 a(n) is the smallest k such that the p-rank of (Z/kZ)* is 2, where p = prime(n) and (Z/kZ)* is the multiplicative group of integers modulo n.

Original entry on oeis.org

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
Offset: 1

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Author

Jianing Song, Apr 08 2019

Keywords

Comments

The p-rank 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 p-rank 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).

Examples

			(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).
		

Crossrefs

Programs

  • 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)))))

Formula

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