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%I #65 Aug 26 2021 21:08:45
%S 2,5,3,29,7,13,17,197,41,5741,11,33461,239,269,577,137,199,37,19,
%T 45697,23,229,1153,1549,79,53,113,44560482149,31,61,665857,52734529,
%U 103,1800193921,73,593,9369319,389,241,1746860020068409,4663,11437,43,6481,47,3761,97,293,45245801,101,22307,68480406462161287469,7761799,109,1535466241
%N a(n) = smallest prime which divides Pell(n) = A000129(n) but does not divide any Pell(k) for k<n, or -1 if no such prime exists.
%C First differs from A264137 (Largest prime factor of the n-th Pell number) at n=17; see Example section. - _Jon E. Schoenfield_, Dec 10 2016
%H <a href="/A246556/b246556.txt">Table of n, a(n) for n = 2..630</a>
%H R. D. Carmichael, <a href="https://doi.org/10.2307%2F1967797">On the numerical factors of the arithmetic forms α^n ± β^n</a>, Annals of Math., 15 (1/4) (1913), 30-70.
%F a(n) >= 2 for all n >= 2, by Carmichael's theorem. - _Jonathan Sondow_, Dec 08 2017
%e a(2) = 2 because Pell(2) = 2 and Pell(k) < 2 for k < 2.
%e a(4) = 3 because Pell(4) = 12 = 2^2 * 3, but 2 is not a primitive prime factor since Pell(2) = 2, so therefore 3 is the primitive prime factor.
%e a(5) = 29 because Pell(5) = 29, which is prime.
%e a(6) = 7 because Pell(6) = 70 = 2 * 5 * 7, but neither 2 nor 5 is a primitive prime factor, so therefore 7 is the primitive prime factor.
%e a(17) = 137 because Pell(17) = 1136689 = 137 * 8297, and both of them are primitive factors, we choose the smallest. (Pell(17) is the smallest Pell number with more than one primitive prime factor.)
%t prms={}; Table[f=First/@FactorInteger[Pell[n]]; p=Complement[f, prms]; prms=Join[prms, p]; If[p=={}, 1, First[p]], {n, 36}]
%Y Cf. A001578 (for Fibonacci(n)), A000129 (Pell numbers), A008555, A086383, A096650, A120947, A175181, A214028, A264137.
%K nonn
%O 2,1
%A _Eric Chen_, Nov 15 2014
%E Edited by _N. J. A. Sloane_, Nov 29 2014
%E Terms up to a(612) in b-file added by _Sean A. Irvine_, Sep 23 2019
%E Terms a(613)-a(630) in b-file added by _Max Alekseyev_, Aug 26 2021
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