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A057781 a(n) = n^4+4 = (n^2-2*n+2)*(n^2+2*n+2) = ((n-1)^2+1)*((n+1)^2+1).

%I A057781 #45 Sep 08 2022 08:45:02
%S A057781 4,5,20,85,260,629,1300,2405,4100,6565,10004,14645,20740,28565,38420,
%T A057781 50629,65540,83525,104980,130325,160004,194485,234260,279845,331780,
%U A057781 390629,456980,531445,614660,707285,810004,923525,1048580,1185925
%N A057781 a(n) = n^4+4 = (n^2-2*n+2)*(n^2+2*n+2) = ((n-1)^2+1)*((n+1)^2+1).
%D A057781 Donald E. Knuth, The Art of Computer Programming, Addison-Wesley, Reading, MA, 1997, Vol. 1, exercise 1.2.1, Nr. 11, p. 19. [From _Reinhard Zumkeller_, Apr 11 2010]
%H A057781 Vincenzo Librandi, <a href="/A057781/b057781.txt">Table of n, a(n) for n = 0..10000</a>
%H A057781 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A057781 G.f.: -(5*x^4-5*x^3+35*x^2-15*x+4) / (x-1)^5. - _Colin Barker_, Mar 29 2013
%F A057781 a(n) = A002523(n) + 3.
%F A057781 a(n) = A002522(n-1) * A002522(n+1).
%F A057781 Sum_{k=0..n} A033999(k)*A016755(k)/a(k) = A033999(n)*(n+1)/A053755(n+1), see Knuth reference. - _Reinhard Zumkeller_, Apr 11 2010
%F A057781 a(n) = (n^2)^2 + 2^2 = (n^2-2)^2 + (2*n)^2. - _Thomas Ordowski_, Sep 15 2015
%F A057781 a(n) = A272298(3*n)/3^4. - _Bruno Berselli_, Apr 29 2016
%F A057781 Sum_{n>=0} 1/a(n) = (Pi*coth(Pi) + 1)/8. - _Amiram Eldar_, Oct 04 2021
%t A057781 Table[n^4+4,{n,0,60}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 15 2011 *)
%t A057781 LinearRecurrence[{5,-10,10,-5,1},{4,5,20,85,260},40] (* _Harvey P. Dale_, Aug 20 2020 *)
%o A057781 (Magma) [n^4+4: n in [0..40]]; // _Vincenzo Librandi_, Sep 07 2011
%o A057781 (PARI) first(m)=vector(m,i,i--;i^4+4) \\ _Anders Hellström_, Sep 15 2015
%Y A057781 Cf. A016755, A033999, A053755.
%Y A057781 Cf. A000583, A002522, A002523, A272298.
%K A057781 nonn,easy
%O A057781 0,1
%A A057781 _Henry Bottomley_, Nov 04 2000