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

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

Author

Rafael Parra Machio, May 19 2011

Keywords

• nonn
• easy

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.

Examples

			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.
		

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).

Programs

Magma

[n^4+2^4+(n+2)^4: n in [0..35]]; // Vincenzo Librandi, Jun 09 2011

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 *)

PARI

a(n)=2*(n^2+2*n+4)^2 \\ Charles R Greathouse IV, Jun 08 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

Formula

G.f.: (32 - 62*x + 118*x^2 - 58*x^3 + 18*x^4)/(1-x)^5.