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 A190176 a(n) = n^4 + 2^4 + (n+2)^4. 2
 32, 98, 288, 722, 1568, 3042, 5408, 8978, 14112, 21218, 30752, 43218, 59168, 79202, 103968, 134162, 170528, 213858, 264992, 324818, 394272, 474338, 566048, 670482, 788768, 922082, 1071648, 1238738, 1424672, 1630818, 1858592 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS 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. More generally, n^4 + k^4 + (n+k)^4 = 2*(n^2 + n*k + k^2)^2; in this case, k=2. REFERENCES Robert Carmichael, Diophantine Analysis, Ed. 1915 by Mathematical Monographs, pages 66-67. LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..3000 Rafael Parra Machío, dofanticas.pdf, pages 14-15 Rafael Parra Machío, Educaciones iofanticas. Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1). FORMULA G.f.: (32 - 62*x + 118*x^2 - 58*x^3 + 18*x^4)/(1-x)^5. EXAMPLE a(3) = 722 = 3^4 +2^4+(3+2)^4 = 2(3^2+3*2+2^2)^2 = 2*19^2. a(13) = 79202 = 13^4+2^4+(13 + 2)^4 = 2(13^2+13*2+2^2)^2 = 2*199^2. MATHEMATICA Table[n^4+2^4+(n+2)^4, {n, 0, 20}] 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 *) PROG (PARI) a(n)=2*(n^2+2*n+4)^2 \\ Charles R Greathouse IV, Jun 08 2011 (MAGMA) [n^4+2^4+(n+2)^4: n in [0..35]]; // Vincenzo Librandi, Jun 09 2011 (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 CROSSREFS Sequence in context: A197604 A287925 A039519 * A198070 A197904 A273554 Adjacent sequences:  A190173 A190174 A190175 * A190177 A190178 A190179 KEYWORD nonn,easy AUTHOR Rafael Parra Machio, May 19 2011 STATUS approved

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Last modified February 18 15:30 EST 2020. Contains 332019 sequences. (Running on oeis4.)