login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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
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
Rafael Parra Machío, dofanticas.pdf, pages 14-15
Rafael Parra Machío, Educaciones iofanticas.
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 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {32, 98, 288, 722, 1568}, 50] (* Harvey P. Dale, May 26 2023 *)
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
KEYWORD
nonn,easy
AUTHOR
Rafael Parra Machio, May 19 2011
STATUS
approved