%I #45 Sep 08 2022 08:45:02

%S 4,5,20,85,260,629,1300,2405,4100,6565,10004,14645,20740,28565,38420,

%T 50629,65540,83525,104980,130325,160004,194485,234260,279845,331780,

%U 390629,456980,531445,614660,707285,810004,923525,1048580,1185925

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

%D 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 Vincenzo Librandi, <a href="/A057781/b057781.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F G.f.: -(5*x^4-5*x^3+35*x^2-15*x+4) / (x-1)^5. - _Colin Barker_, Mar 29 2013

%F a(n) = A002523(n) + 3.

%F a(n) = A002522(n-1) * A002522(n+1).

%F 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 a(n) = (n^2)^2 + 2^2 = (n^2-2)^2 + (2*n)^2. - _Thomas Ordowski_, Sep 15 2015

%F a(n) = A272298(3*n)/3^4. - _Bruno Berselli_, Apr 29 2016

%F Sum_{n>=0} 1/a(n) = (Pi*coth(Pi) + 1)/8. - _Amiram Eldar_, Oct 04 2021

%t Table[n^4+4,{n,0,60}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 15 2011 *)

%t LinearRecurrence[{5,-10,10,-5,1},{4,5,20,85,260},40] (* _Harvey P. Dale_, Aug 20 2020 *)

%o (Magma) [n^4+4: n in [0..40]]; // _Vincenzo Librandi_, Sep 07 2011

%o (PARI) first(m)=vector(m,i,i--;i^4+4) \\ _Anders Hellström_, Sep 15 2015

%Y Cf. A016755, A033999, A053755.

%Y Cf. A000583, A002522, A002523, A272298.

%K nonn,easy

%O 0,1

%A _Henry Bottomley_, Nov 04 2000