OFFSET
1,1
COMMENTS
If k is such number, then b^p == b^q (mod k) for every integer b.
Problem: are there infinitely many such numbers?
Suppose p^2 divides k. Then p divides k - phi(k), and so the only way k - phi(k) can be prime is if k = p^2. But then k - phi(k) = k - A002322(k). Hence all terms in this sequence are squarefree. - Charles R Greathouse IV, Oct 08 2016
All terms are odd composites. - Robert Israel, Oct 09 2016
It seems that gpf(k) < p = k - phi(k). - Thomas Ordowski, Oct 09 2016
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
EXAMPLE
MAPLE
filter:= proc(n) uses numtheory;
local p, q;
p:= n-phi(n);
q:= n-lambda(n);
p<q and isprime(p) and isprime(q);
end proc:
select(filter, [seq(i, i=3..10000, 2)]); # Robert Israel, Oct 09 2016
MATHEMATICA
Select[Range[10^3], And[#1 < #2, Times @@ Boole@ PrimeQ@ {#1, #2} == 1] & @@ {# - EulerPhi@ #, # - CarmichaelLambda@ #} &] (* Michael De Vlieger, Oct 08 2016 *)
PROG
(PARI) is(n)=my(f=factor(n), p=n-eulerphi(f), q=n-lcm(znstar(f)[2])); p < q && isprime(p) && isprime(q) \\ Charles R Greathouse IV, Oct 08 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Thomas Ordowski, Oct 07 2016
EXTENSIONS
More terms from Altug Alkan, Oct 07 2016
STATUS
approved
