

A307437


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


2



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
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OFFSET

1,1


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, ...


LINKS

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


EXAMPLE

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.


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


PROG

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


CROSSREFS

Cf. A002322, A307436.
Sequence in context: A010070 A291107 A132445 * A070846 A078683 A099863
Adjacent sequences: A307434 A307435 A307436 * A307438 A307439 A307440


KEYWORD

nonn,look


AUTHOR

Jianing Song, Apr 08 2019


STATUS

approved



