|
%I #24 Jun 08 2022 10:16:39
%S 5,13,41,113,109,2113,1613,1321,26317,525313,14449,30269,268501,
%T 279073,536903681,384773,4327489,47392381,231769777,21841,43249589,
%U 1759217765581,29247661,140737471578113,4981857697937,1326700741,1801439824104653,3630105520141
%N Largest prime factor of 4^(2*n+1)+1.
%C 4^(2*n+1)+1 = 2^(2*(2*n+1))+1 = (2^(2*n+1)-2^(n+1)+1) * (2^(2*n+1)+2^(n+1)+1).
%C For all n, the smallest prime factor of 4^(2*n+1)+1 is 5.
%C Therefore, the present sequence also gives the largest prime factor of (4^(2*n+1)+1)/5 = A299960(n), for all n > 0. See A299959 for the smallest prime factor of this. - _M. F. Hasler_, Feb 27 2018
%H Daniel Suteu, <a href="/A229747/b229747.txt">Table of n, a(n) for n = 0..547</a>
%F a(n) = A006530(A052539(2n+1)) = A006530(A207262(n+1)), and for n > 1, a(n) = A006530(A299960(n)) = A006530(A052539(2n+1)/5). \\ _M. F. Hasler_, Feb 27 2018
%F a(n) = max(A229767(n), A229768(n)), for n >= 1. - _Daniel Suteu_, Jun 08 2022
%e For n=7, 4^(2*n+1)+1 = 1073741825 = 5*5*13*41*61*1321. So a(7)=1321.
%t Table[FactorInteger[4^(2n+1)+1][[-1,1]],{n,0,30}] (* _Harvey P. Dale_, Mar 10 2018 *)
%o (PARI) a(n) = {
%o f=factor(2^(2*n+1)-2^(n+1)+1);
%o g=factor(2^(2*n+1)+2^(n+1)+1);
%o max(f[matsize(f)[1],1], g[matsize(g)[1],1])
%o }
%Y Cf. A207262. Bisection of A274903.
%Y Cf. A299960, A299959, A052539.
%K nonn
%O 0,1
%A _Colin Barker_, Sep 28 2013
|
