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A292538
Lucas-Carmichael 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
OFFSET
1,1
COMMENTS
Intersection of A005563 and A006972.
The numbers k such that k^2 - 1 is a Lucas-Carmichael number are 20, 54, 84, 252, 384, 1260, 1308, 1500, 1710, 2580, 3080, 3780, 4392, ...
From David A. Corneth, Aug 26 2023: (Start)
As k^2 - 1 = (k - 1)*(k + 1) and k is even we have k-1 and k+1 are coprime. So we can factor k-1 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
David A. Corneth, Table of n, a(n) for n = 1..2391 (terms <= 4*10^17; first 164 terms from Amiram Eldar)
MAPLE
filter:= t ->
andmap(f -> f[2]=1 and (t+1) mod (f[1]+1) = 0, ifactors(t)[2]):
select(filter, [seq(k^2-1, 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
Sequence in context: A292573 A299213 A206536 * A369246 A065767 A166915
KEYWORD
nonn
AUTHOR
Amiram Eldar, Sep 18 2017
EXTENSIONS
More terms from David A. Corneth, Aug 26 2023
STATUS
approved