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A318688
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Numbers k such that 2^(3^j) == 1 (mod k) for some j.
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1
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1, 7, 73, 487, 511, 2593, 3409, 18151, 35551, 39367, 71119, 80191, 97687, 189289, 209953, 248857, 262657, 275569, 379081, 472393, 497833, 561337, 683809, 1262791, 1299079, 1325023, 1469671, 1838599, 2653567, 2873791, 3306751, 5191687, 5853943, 7131151, 8839537, 9093553, 15326569, 19171729
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OFFSET
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1,2
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COMMENTS
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Numbers k such that the multiplicative order of 2 mod k is a power of 3.
If x and y are coprime members of the sequence, then x*y is in the sequence.
All divisors of a member of the sequence are in the sequence.
All prime-power divisors of 2^(3^k)-1 are in the sequence. In particular, the sequence contains infinitely many primes. - Robert Israel, Sep 02 2018
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LINKS
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EXAMPLE
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a(3) = 73 is in the sequence because the multiplicative order of 2 mod 73 is 9 which is a power of 3.
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MAPLE
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N:= 10^6: # to get all terms <= N
Res:= NULL:
p:= 5:
do
p:= nextprime(p);
if p > N then break fi;
q:= 1: t:= 2:
while q < p-1 do
q:= 3*q;
t:= t^3 mod p;
if t = 1 then
Res:= Res, p;
v:= 1;
while 2 &^ t mod (p^(v+1)) = 1 do v:= v+1 od:
V[p]:= v;
break
fi
od
od:
S:= {1}:
for p in Res do
S:= `union`(S, seq(map(`*`, select(`<=`, S, N/p^i), p^i), i=1..V[p]))
od:
sort(convert(S, list));
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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