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A299959
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Least prime factor of (4^(2n+1)+1)/5, a(0) = 1.
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3
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1, 13, 5, 29, 13, 397, 53, 5, 137, 229, 13, 277, 5, 13, 107367629, 5581, 13, 5, 149, 13, 10169, 173, 5, 3761, 29, 13, 15358129, 5, 13, 1181, 733, 13, 5, 269, 13, 569, 293, 5, 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
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OFFSET
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0,2
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LINKS
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FORMULA
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a(n) = 5 iff n = 2 (mod 5); otherwise, a(n) = 13 if n = 1 (mod 3).
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EXAMPLE
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For n = 0, A299960(0) = (4^1+1)/5 = 5/5 = 1, therefore we let a(0) = 1.
For n = 1, A299960(1) = (4^3+1)/5 = 65/5 = 13 is prime, therefore a(1) = 13.
For n = 2, A299960(2) = (4^5+1)/5 = 1025/5 = 205 = 5*41, therefore a(2) = 5.
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PROG
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(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.
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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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