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A229747
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Largest prime factor of 4^(2*n+1)+1.
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4
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5, 13, 41, 113, 109, 2113, 1613, 1321, 26317, 525313, 14449, 30269, 268501, 279073, 536903681, 384773, 4327489, 47392381, 231769777, 21841, 43249589, 1759217765581, 29247661, 140737471578113, 4981857697937, 1326700741, 1801439824104653, 3630105520141
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OFFSET
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0,1
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COMMENTS
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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).
For all n, the smallest prime factor of 4^(2*n+1)+1 is 5.
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
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LINKS
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FORMULA
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EXAMPLE
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For n=7, 4^(2*n+1)+1 = 1073741825 = 5*5*13*41*61*1321. So a(7)=1321.
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MATHEMATICA
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Table[FactorInteger[4^(2n+1)+1][[-1, 1]], {n, 0, 30}] (* Harvey P. Dale, Mar 10 2018 *)
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PROG
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(PARI) a(n) = {
f=factor(2^(2*n+1)-2^(n+1)+1);
g=factor(2^(2*n+1)+2^(n+1)+1);
max(f[matsize(f)[1], 1], g[matsize(g)[1], 1])
}
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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