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Numbers k such that lambda(k)^3 divides (k-1)^2, where lambda(k) = A002322(k).
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%I #45 Apr 21 2024 09:59:36

%S 1,2,1729,19683001,367804801,631071001,2064236401,2320690177,

%T 24899816449,40017045601,110592000001,137299665601,432081216001,

%U 479534887801,760355883001,1111195454401,3176523000001,3495866888449,3837165696001,8571867768001,14373832968001

%N Numbers k such that lambda(k)^3 divides (k-1)^2, where lambda(k) = A002322(k).

%C Carmichael numbers are composite numbers n such that k = 1 (mod lambda(k)); equivalently, lambda(k)^2 divides (k-1)^2. As a result, all composite terms of the sequence are Carmichael numbers A002997. But there are no primes in this sequence except for 2 (since lambda(p) = p-1 and (p-1)^3 > (p-1)^2 for p > 2) and so all terms in this sequence other than 1 and 2 are Carmichael numbers. - _Charles R Greathouse IV_, Oct 15 2016

%C Is this sequence infinite?

%H Amiram Eldar, <a href="/A277389/b277389.txt">Table of n, a(n) for n = 1..1469</a> (terms below 10^22, calculated using data from Claude Goutier; terms 1..58 from Robert Israel, terms 59..101 from Charles R Greathouse IV)

%H Claude Goutier, <a href="http://www-labs.iro.umontreal.ca/~goutier/OEIS/A055553/">Compressed text file carm10e22.gz containing all the Carmichael numbers up to 10^22</a>.

%o (PARI) isok(n) = ((n-1)^2 % (lcm(znstar(n)[2])^3)) == 0; \\ _Michel Marcus_, Oct 12 2016

%Y Cf. A002322, A002997, A265628.

%K nonn

%O 1,2

%A _Thomas Ordowski_, Oct 12 2016

%E a(4) from _Michel Marcus_, Oct 12 2016

%E a(5)-a(6) from _Michel Marcus_, Oct 13 2016

%E More terms from _Robert Israel_, Oct 13 2016