OFFSET
1,1
COMMENTS
This sequence is the complement of A197504 with respect to the positive odd numbers. It collects all positive odd numbers m with 4 dividing phi(m).
In the prime number factorization of an odd m >= 3 the largest even factor of phi(a(n)) is 2^{2*r1 + r3}, where r1 and r3 are the nonnegative number of distinct primes 1 (mod 4) and 3 (mod 4), respectively. This means that for m = a(n) one needs 2*r1 + r3 >= 2. See some examples below.
The number of solutions of the congruence x^2 == +1 (mod a(n)) or (inclusive) x^2 == -1 (mod a(n)) is 2^(r1 + r3) + delta_{r3,0}*2^r1, with 2*r1 + r3 >= 2, where r1 and r3 are the number of distinct primes 1 (mod 4) and 3 (mod 4), respectively, in the prime number factorization of a(n), and delta is the Kronecker symbol.
This follows from the result that primes 1 (mod 4) (A002144) have Legendre symbol (-1, p) = +1 and primes 3 (mod 4) (A002145) have (-1, p) = -1. The part (a) of the lifting theorem for powers of primes (Apostol, 5.30, pp. 121-122) is used. Also the theorem that for odd primes p there are exactly (p-1)/2 quadratic residues modulo p (and exactly (p-1)/2 nonresidues modulo p) is needed (see e.g., Silverman, ch. 20, Theorem 1, p. 151).
REFERENCES
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, pp. 121-122.
J. H. Silverman, A Friendly Introduction to Number Theory, fourth ed., Pearson Education, Inc, 2014, ch. 20, pp. 149-155.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..10000
FORMULA
All members of the set {odd m >= 1: 4 | phi(m)} ordered increasingly.
EXAMPLE
[n, a(n), [r1, r3], number of solutions x (mod a(n)), [solutions]] (with pm for + or -):
[1, 5, [1, 0], 4, [pm1,pm2]],
[5, 21 = 3*7, [0, 2], 4, [pm1, pm8],
[20, 65 = 5*13, [1, 1], 8, [pm1, pm8, pm14, pm18]],
[34, 105 = 3*5*7, [1, 2], 8, [pm1, pm29, pm34, pm41].
MATHEMATICA
Select[Range[3, 200, 2], And[EvenQ[#1], #1 == #2] & @@ {EulerPhi[2 #]/2, EulerPhi[#]/2} &] (* Michael De Vlieger, Jun 28 2020 *)
PROG
(PARI) isok(m) = (m > 3) && (m % 2) && ((eulerphi(m) % 4) == 0); \\ Michel Marcus, Nov 13 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Nov 12 2019
STATUS
approved