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A002523
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a(n) = n^4 + 1.
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34
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1, 2, 17, 82, 257, 626, 1297, 2402, 4097, 6562, 10001, 14642, 20737, 28562, 38417, 50626, 65537, 83522, 104977, 130322, 160001, 194482, 234257, 279842, 331777, 390626, 456977, 531442, 614657, 707282, 810001, 923522, 1048577, 1185922, 1336337, 1500626, 1679617
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OFFSET
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0,2
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COMMENTS
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a(n) = Phi_8(n), where Phi_k is the k-th cyclotomic polynomial.
All odd prime factors of a(n) are congruent to 1 modulo 8. - Nick Hobson, Jan 14 2007
Lee and Murty, p. 685: "In spite of these remarkable advances, we are still unable to determine if n^4 + 1 is infinitely often a squarefree number". - Jonathan Vos Post, Sep 18 2007
Since a(n)*a(m) = (n^4+1)*(m^4+1) = ((n*m)^2-1)^2 + (n^2+m^2)^2, a(n)*a(m) is obvious member of A000404 for n*m > 1. Additionally, if m and n are the legs of a Pythagorean triple, then a(m)*a(n) is the member of A111925. - Altug Alkan, Apr 08 2016
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REFERENCES
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M. Mabkhout, "Minoration de P(x^4+1)", Rend. Sem. Fac. Sci. Univ. Cagliari 63 (2) (1993), 135-148.
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LINKS
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FORMULA
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O.g.f.: (1 - 3*x + 17*x^2 + 7*x^3 + 2*x^4)/(1-x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
Sum_{n>=0} 1/a(n) = 1/2 + Pi * (sinh(sqrt(2)*Pi) + sin(sqrt(2)*Pi)) / (2*sqrt(2) * (cosh(sqrt(2)*Pi) - cos(sqrt(2)*Pi))) = 1.578477579667136838318... . - Vaclav Kotesovec, Feb 14 2015
Sum_{n>=0} (-1)^n/a(n) = 1/2 - Pi * (cos(Pi/sqrt(2)) * sinh(Pi/sqrt(2)) + cosh(Pi/sqrt(2)) * sin(Pi/sqrt(2))) / (sqrt(2) * (cos(sqrt(2)*Pi) - cosh(sqrt(2)*Pi))) = 0.54942814871987317922929... . - Vaclav Kotesovec, Feb 14 2015
Product_{n>=1} (1 - 1/a(n)) = 2*Pi^2/(cosh(sqrt(2)*Pi) - cos(sqrt(2)*Pi)). - Amiram Eldar, Jan 26 2024
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MAPLE
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numtheory[cyclotomic](8, n) ;
end proc:
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MATHEMATICA
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LinearRecurrence[{5, -10, 10, -5, 1}, {1, 2, 17, 82, 257}, 30] (* Ray Chandler, Aug 26 2015 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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