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A299959 Least prime factor of (4^(2n+1)+1)/5, a(0) = 1.

%I A299959 #20 Jan 13 2025 11:42:21
%S A299959 1,13,5,29,13,397,53,5,137,229,13,277,5,13,107367629,5581,13,5,149,13,
%T A299959 10169,173,5,3761,29,13,15358129,5,13,1181,733,13,5,269,13,569,293,5,
%U A299959 29,317,13,997,5,13,1069,29,13,5,389,13,809,41201,5,857,5669,13,58309,5,13,29,397,13,5,509,13
%N A299959 Least prime factor of (4^(2n+1)+1)/5, a(0) = 1.
%C A299959 The range of this sequence with a(0) = 1 omitted, {5, 13, 29, 53, 137, ...}, appears to be a subset of A261580 (and of the Pythagorean primes A002144). Is there a smaller superset sequence in OEIS? - _M. F. Hasler_, Jan 07 2025
%H A299959 Chai Wah Wu, <a href="/A299959/b299959.txt">Table of n, a(n) for n = 0..715</a>
%F A299959 a(n) = A020639(A299960(n)) = A020639(A052539(2n+1)/5).
%F A299959 a(n) = 5 iff n = 2 (mod 5); otherwise, a(n) = 13 if n = 1 (mod 3).
%F A299959 Otherwise, a(n) = 29 if n = 3 (mod 7), else a(n) = 53 if n = 6 (mod 13), else a(n) = 137 if n = 8 (mod 17), else a(n) = 149 if n = 18 (mod 27), else a(n) = 173 if n = 21 (mod 43), etc... - _M. F. Hasler_, Jan 07 2025
%e A299959 For n = 0, A299960(0) = (4^1+1)/5 = 5/5 = 1, therefore we let a(0) = 1.
%e A299959 For n = 1, A299960(1) = (4^3+1)/5 = 65/5 = 13 is prime, therefore a(1) = 13.
%e A299959 For n = 2, A299960(2) = (4^5+1)/5 = 1025/5 = 205 = 5*41, therefore a(2) = 5.
%o A299959 (PARI) a(n)=A020639(4^(2*n+1)\5+1) \\ Using factor(...)[1,1] requires complete factorization and is much less efficient for large n.
%Y A299959 Cf. A299960, A052539.
%K A299959 nonn
%O A299959 0,2
%A A299959 _M. F. Hasler_, Feb 22 2018