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A190176 a(n) = n^4 + 2^4 + (n+2)^4.

%I A190176 #35 May 26 2023 16:45:26
%S A190176 32,98,288,722,1568,3042,5408,8978,14112,21218,30752,43218,59168,
%T A190176 79202,103968,134162,170528,213858,264992,324818,394272,474338,566048,
%U A190176 670482,788768,922082,1071648,1238738,1424672,1630818,1858592
%N A190176 a(n) = n^4 + 2^4 + (n+2)^4.
%C A190176 Each term equals the sum of three fourth powers and also twice a perfect square: n^4 + 2^4 + (n+2)^4 = 2*(n^2 + 2*n + 2^2)^2.
%C A190176 More generally, n^4 + k^4 + (n+k)^4 = 2*(n^2 + n*k + k^2)^2; in this case, k=2.
%D A190176 Robert Carmichael, Diophantine Analysis, Ed. 1915 by Mathematical Monographs, pages 66-67.
%H A190176 Vincenzo Librandi, <a href="/A190176/b190176.txt">Table of n, a(n) for n = 0..3000</a>
%H A190176 Rafael Parra Machío, <a href="http://hojamat.es/parra/diofanticas.pdf">dofanticas.pdf</a>, pages 14-15
%H A190176 Rafael Parra Machío, <a href="http://hojamat.es/parra/iniparra.htm">Educaciones iofanticas</a>.
%H A190176 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A190176 G.f.: (32 - 62*x + 118*x^2 - 58*x^3 + 18*x^4)/(1-x)^5.
%e A190176 a(3) = 722 = 3^4 +2^4+(3+2)^4 = 2(3^2+3*2+2^2)^2 = 2*19^2.
%e A190176 a(13) = 79202 = 13^4+2^4+(13 + 2)^4 = 2(13^2+13*2+2^2)^2 = 2*199^2.
%t A190176 Table[n^4+2^4+(n+2)^4,{n,0,20}]
%t A190176 CoefficientList[Series[(32 - 62*x + 118*x^2 - 58*x^3 + 18*x^4)/(1-x)^5, {x,0,50}], x] (* _G. C. Greubel_, Dec 28 2017 *)
%t A190176 LinearRecurrence[{5,-10,10,-5,1},{32,98,288,722,1568},50] (* _Harvey P. Dale_, May 26 2023 *)
%o A190176 (PARI) a(n)=2*(n^2+2*n+4)^2 \\ _Charles R Greathouse IV_, Jun 08 2011
%o A190176 (Magma) [n^4+2^4+(n+2)^4: n in [0..35]]; // _Vincenzo Librandi_, Jun 09 2011
%o A190176 (PARI) x='x+O('x^30); Vec((32 - 62*x + 118*x^2 - 58*x^3 + 18*x^4)/(1-x)^5 ) \\ _G. C. Greubel_, Dec 28 2017
%K A190176 nonn,easy
%O A190176 0,1
%A A190176 _Rafael Parra Machio_, May 19 2011