

A292538


LucasCarmichael numbers of the form k^2  1.


2



399, 2915, 7055, 63503, 147455, 1587599, 1710863, 2249999, 2924099, 6656399, 9486399, 14288399, 19289663, 25603599, 26936099, 28451555, 31270463, 32148899, 45158399, 49280399, 71368703, 91011599, 105884099, 111513599, 144288143, 146894399, 150405695, 152028899, 175827599
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OFFSET

1,1


COMMENTS

The numbers k such that k^2  1 is a LucasCarmichael number are 20, 54, 84, 252, 384, 1260, 1308, 1500, 1710, 2580, 3080, 3780, 4392, ...
As k^2  1 = (k  1)*(k + 1) and k is even we have k1 and k+1 are coprime. So we can factor k1 and k+1 separately when checking if k^2  1 is a term.
Possible other ideas are factoring an odd number only once, keeping it for the factorization of k^2  1 and (k + 2)^2  1. Alternatively dodging k = 18m + 8, 18m + 10 or 50m + 24, 50m + 26 to not get numbers that are multiples of odd primes squared. (End)


LINKS



MAPLE

filter:= t >
andmap(f > f[2]=1 and (t+1) mod (f[1]+1) = 0, ifactors(t)[2]):
select(filter, [seq(k^21, k=3..10^5)]); # Robert Israel, Sep 24 2017


MATHEMATICA

lcQ[n_] := !PrimeQ[n] && Union[Transpose[FactorInteger[n]][[2]]] == {1} && Union[Mod[n + 1, Transpose[FactorInteger[n]][[1]] + 1]] == {0}; Select[Range[2, 10^4]^2  1, lcQ]


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



