A307437 a(n) is the smallest k such that 2n divides psi(k), psi = A002322.

3, 5, 7, 17, 11, 13, 29, 17, 19, 25, 23, 73, 53, 29, 31, 97, 103, 37, 191, 41, 43, 89, 47, 97, 101, 53, 81, 113, 59, 61, 311, 193, 67, 137, 71, 73, 149, 229, 79, 187, 83, 203, 173, 89, 181, 235, 283, 97, 197, 101, 103, 313, 107, 109, 121, 113, 229, 233, 709, 241

Offset: 1

Author

Jianing Song, Apr 08 2019

Keywords

• nonn
• look

Comments

a(n) exists for all n: by Dirichlet's theorem on arithmetic progressions, there exists a prime p congruent to 1 modulo 2n, so 2n divides psi(p) = p - 1.

a(n) is the smallest k such that (Z/kZ)* contains C_(2n) as a subgroup, where (Z/kZ)* is the multiplicative group of integers modulo n.

a(n) is the smallest k such that there exists some x such that ord(x,k) = 2n, where ord(x,k) is the multiplicative order of x modulo k.

Record values of a(n)/n occur at n = 1, 4, 12, 19, 59, 167, 196, 197, 227, 317, 457, 521, 706, ... (A341888).

From Jianing Song, Feb 21 2021: (Start)

a(n) is bounded above by (2n)^2 since n divides psi(n^2).

a(n) is usually odd. There are only 7 values <= 10^4 for n such that a(n) is even, namely n = 256, 512, 1024, 2816, 4096, 5632 and 8192 (A341887). (End)

a(n) is odd or divisible by 16, since psi(k) = psi(2k) = psi(4k) = psi(8k) for odd k > 1. - Jianing Song, Feb 22 2021

Examples

			For n = 7, psi(29) = 28 and 29 is the smallest k such that 14 divides psi(k), so a(7) = 29.
For n = 27, psi(81) = 54 and 81 is the smallest k such that 54 divides psi(k), so a(27) = 81.
For n = 40, psi(187) = 80 and 187 is the smallest k such that 80 divides psi(k), so a(40) = 187.
For n = 42, psi(203) = 84 and 203 is the smallest k such that 84 divides psi(k), so a(42) = 203.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Wikipedia, Multiplicative group of integers modulo n

Crossrefs

Cf. A002322, A307436, A341887, A341888.

Programs

Maple

N:= 100: # for a(1)..a(N)
V:= Vector(N): count:= 0:
for k from 3 while count < N do
  S:= select(t -> t <= N and V[t]=0, numtheory:-divisors(numtheory:-lambda(k)/2));
  if nops(S) > 0 then count:= count + nops(S); V[convert(S,list)]:= k fi
od:
convert(V,list); # Robert Israel, Jul 10 2019

PARI

a(n) = my(i=1); while(A002322(i)%(2*n), i++); i \\ See A002322 for its program

Python

from sympy import reduced_totient
def A307437(n):
    k = 1
    while reduced_totient(k) % (2*n):
        k += 1
    return k # Chai Wah Wu, Feb 24 2021

Formula

From Jianing Song, Feb 26 2021: (Start)

For odd prime p, a((p-1)/2*p^e) = p^(e+1) if (p-1)*p^e+1 is composite, (p-1)*p^e+1 otherwise. Proof: suppose a((p-1)/2*p^e) = p^a*r < p^(e+1), p does not divide r, then (p-1)*p^e | lcm((p-1)*p^(a-1), psi(r)) => p^e | lcm(p^(a-1), psi(r)).

If p^e | p^(a-1), then a((p-1)/2*p^e) >= p^a >= p^(e+1).

If p^e does not divide p^(a-1), then p^e | psi(r). r must have a prime factor of the form q = 2*t*p^e+1. If a >= 1, then a((p-1)/2*p^e) >= p*(2*p^e+1) > p^(e+1). So we must have a = 0. Write r = r'*q^b, then p-1 | lcm(psi(r'), 2*t*p^e*q^(b-1)) => p-1 | lcm(psi(r'), 2*t), hence 2*t*r' >= 2*t*psi(r') >= lcm(psi(r'), 2*t) => p-1. If 2*t*r' > p-1, then a((p-1)/2*p^e) >= r'*q = r'*(2*t*p^e+1) > p^(e+1). If 2*t*r' = p-1, then r' = psi(r') => r' = 1, 2*t = p-1, hence (p-1)*p^e+1 is prime. (End)