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A211412
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a(n) = 4*n^4 + 1.
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4
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5, 65, 325, 1025, 2501, 5185, 9605, 16385, 26245, 40001, 58565, 82945, 114245, 153665, 202501, 262145, 334085, 419905, 521285, 640001, 777925, 937025, 1119365, 1327105, 1562501, 1827905, 2125765, 2458625, 2829125, 3240001, 3694085, 4194305, 4743685, 5345345, 6002501, 6718465, 7496645
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OFFSET
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1,1
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COMMENTS
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Except for the first term, all terms are composite. a(n) is divisible by 5 if n is not.
Long before Aurifeuille, Euler discovered that 4n^4 + 1 = (2n^2 + 2n + 1)*(2n^2 - 2n + 1). For example, 325 = 4 * 3^4 + 1 = (2 * 3^2 + 2 * 3 + 1)*(2 * 3^2 - 2 * 3 + 1) = 25 * 13. Euler shared this discovery with Goldbach in a letter dated August 28, 1742. [Euler identity corrected by Graham Holmes, Jun 02 2023]
The terms of the sequence are the arithmetic mean of eight numbers located on concentric circles (see Avilov link). - Nicolay Avilov, Jan 22 2021
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REFERENCES
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Don Knuth, The Art of Computer Programming: Seminumerical Algorithms, 3rd ed., New York: Addison-Wesley Professional (1997), p. 392.
David Wells, Prime Numbers: The Most Mysterious Figures in Math. Hoboken, New Jersey: John Wiley & Sons (2005), p. 15.
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LINKS
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FORMULA
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G.f.: -x*(x^4+50*x^2+40*x+5) / (x-1)^5. - Colin Barker, Feb 11 2013
a(n) = (2*n^2)^2 + 1^2 = (2*n^2-1)^2 + (2*n)^2. - Thomas Ordowski, Sep 18 2015
Sum_{n>=1} 1/a(n) = tanh(Pi/2)*Pi/4 - 1/2.
Sum_{n>=1} (-1)^n/a(n) = 1/2 - sech(Pi/2)*Pi/4. (End)
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MATHEMATICA
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4 Range[44]^4 + 1
LinearRecurrence[{5, -10, 10, -5, 1}, {5, 65, 325, 1025, 2501}, 40] (* Harvey P. Dale, Sep 15 2023 *)
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PROG
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CROSSREFS
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After the first term, subsequence of A121944.
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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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