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%I #36 Sep 08 2022 08:45:03
%S 1,1,13,73,241,601,1261,2353,4033,6481,9901,14521,20593,28393,38221,
%T 50401,65281,83233,104653,129961,159601,194041,233773,279313,331201,
%U 390001,456301,530713,613873,706441,809101,922561,1047553,1184833,1335181,1499401
%N a(n) = n^4 - n^2 + 1.
%C 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
%C 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
%H Harry J. Smith, <a href="/A060886/b060886.txt">Table of n, a(n) for n = 0..1000</a>
%H John Elias, <a href="/A060886/a060886.png">Illustration of initial terms: chain-linked squares</a>
%H <a href="/index/Cy#CyclotomicPolynomialsValuesAtX">Index to values of cyclotomic polynomials of integer argument</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) = Phi_12(n), where Phi_k is the k-th cyclotomic polynomial.
%F G.f.: (1-4*x+18*x^2+8*x^3+x^4)/(1-x)^5. - _Colin Barker_, Apr 21 2012
%F a(n) = (n^2 - 1/2)^2 + 3/4. - _Alonso del Arte_, Dec 20 2015
%F 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
%p A060886 := proc(n)
%p numtheory[cyclotomic](12,n) ;
%p end proc:
%p seq(A060886(n),n=0..20) ; # _R. J. Mathar_, Feb 07 2014
%t (Range[0, 29]^2 - 1/2)^2 + 3/4 (* _Alonso del Arte_, Dec 20 2015 *)
%t Table[n^4 - n^2 + 1, {n, 0, 25}] (* _Vincenzo Librandi_, Dec 20 2015 *)
%o (PARI) a(n) = n^4 - n^2 + 1); \\ _Harry J. Smith_, Jul 14 2009
%o (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
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_, May 05 2001
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