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A280199
Numbers n such that a^(n-1) == 1 (mod n^2) has solutions with 1 < a < n^2-1.
3
5, 7, 11, 13, 15, 17, 19, 21, 23, 25, 28, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 52, 53, 55, 57, 59, 61, 63, 65, 66, 67, 69, 70, 71, 73, 75, 76, 77, 79, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 112, 113, 115, 117, 119, 121, 123, 124, 125
OFFSET
1,1
COMMENTS
Numbers n such that A185103(n) < n^2 + (-1)^n.
Complement of A280196.
Even terms are A039772.
Odd terms are all odd numbers that are not powers of 3.
Conjecture: composite terms are A181780.
LINKS
EXAMPLE
a(4) = 13 is in the sequence because 19^12 == 1 (mod 13^2), and 1 < 19 < 13^2-1.
MAPLE
Aeven:= remove(t -> igcd(t-1, numtheory:-phi(t^2))=1, {seq(i, i=2..1000, 2)}):
Aodd:= {seq(i, i=3..1000, 2)} minus {seq(3^i, i=0..floor(log[3](1000)))}:
sort(convert(Aeven union Aodd, list));
MATHEMATICA
Aeven = DeleteCases[Range[2, 1000, 2], t_ /; GCD[t-1, EulerPhi[t^2]] == 1];
Aodd = Complement[Range[3, 1000, 2], Table[3^i, {i, 0, Floor[Log[3, 1000]]} ]];
Union[Aeven, Aodd] (* Jean-François Alcover, Apr 24 2019, after Robert Israel *)
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved