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%I #28 Nov 06 2023 03:01:06
%S 399,2915,7055,63503,147455,1587599,1710863,2249999,2924099,6656399,
%T 9486399,14288399,19289663,25603599,26936099,28451555,31270463,
%U 32148899,45158399,49280399,71368703,91011599,105884099,111513599,144288143,146894399,150405695,152028899,175827599
%N Lucas-Carmichael numbers of the form k^2 - 1.
%C Intersection of A005563 and A006972.
%C 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, ...
%C From _David A. Corneth_, Aug 26 2023: (Start)
%C 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.
%C 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)
%H David A. Corneth, <a href="/A292538/b292538.txt">Table of n, a(n) for n = 1..2391</a> (terms <= 4*10^17; first 164 terms from Amiram Eldar)
%p filter:= t ->
%p andmap(f -> f[2]=1 and (t+1) mod (f[1]+1) = 0, ifactors(t)[2]):
%p select(filter, [seq(k^2-1, k=3..10^5)]); # _Robert Israel_, Sep 24 2017
%t 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]
%Y Cf. A005563, A006972.
%K nonn
%O 1,1
%A _Amiram Eldar_, Sep 18 2017
%E More terms from _David A. Corneth_, Aug 26 2023
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