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A060886
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a(n) = n^4 - n^2 + 1.
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6
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1, 1, 13, 73, 241, 601, 1261, 2353, 4033, 6481, 9901, 14521, 20593, 28393, 38221, 50401, 65281, 83233, 104653, 129961, 159601, 194041, 233773, 279313, 331201, 390001, 456301, 530713, 613873, 706441, 809101, 922561, 1047553, 1184833, 1335181, 1499401
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OFFSET
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0,3
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COMMENTS
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All positive divisors of a(n) are congruent to 1, modulo 12. Proof: If p is an odd prime different from 3 then n^4 - n^2 + 1 = 0 (mod p) implies: (a) (2n^2 - 1)^2 = -3 (mod p), whence p = 1 (mod 6); and (b) (n^2 - 1)^2 = -n^2 (mod p), whence p = 1 (mod 4). - Nick Hobson, Nov 13 2006
Appears to be the number of distinct possible sums of a set of n distinct integers between 1 and n^3. Checked up to n = 4. - Dylan Hamilton, Sep 21 2010
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LINKS
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FORMULA
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a(n) = Phi_12(n), where Phi_k is the k-th cyclotomic polynomial.
G.f.: (1-4*x+18*x^2+8*x^3+x^4)/(1-x)^5. - Colin Barker, Apr 21 2012
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5), for n>4. - Vincenzo Librandi, Dec 20 2015
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MAPLE
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numtheory[cyclotomic](12, n) ;
end proc:
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MATHEMATICA
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PROG
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(Magma) [n^4 - n^2 + 1: n in [0..40]]; /* or */ I:=[1, 1, 13, 73, 241]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..40]]; // Vincenzo Librandi, Dec 20 2015
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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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