cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A354968 Triangle read by rows: T(n, k) = n*k*(n+k)*(n-k)/6.

Original entry on oeis.org

1, 4, 5, 10, 16, 14, 20, 35, 40, 30, 35, 64, 81, 80, 55, 56, 105, 140, 154, 140, 91, 84, 160, 220, 256, 260, 224, 140, 120, 231, 324, 390, 420, 405, 336, 204, 165, 320, 455, 560, 625, 640, 595, 480, 285, 220, 429, 616, 770, 880, 935, 924, 836, 660, 385, 286, 560, 810, 1024
Offset: 2

Views

Author

Ali Sada and Yifan Xie, Jun 14 2022

Keywords

Comments

Given a Pythagorean triple (a,b,c), define S = c^4 - a^4 - b^4. Using Euclid's parameterization (a = 2*n*k, b = n^2 - k^2, c = n^2 + k^2), substituting to get S in terms of n and k gives S = 8*n^2*k^2*((n^2 - k^2))^2, which is a multiple of 288; T(n, k) = sqrt(S/288) = n*k*(n^2 - k^2)/6 = n*k*(n+k)*(n-k)/6.

Examples

			Triangle begins:
  n/k   1    2    3    4    5    6    7
  2     1;
  3     4,   5;
  4    10,  16,  14;
  5    20,  35,  40,  30;
  6    35,  64,  81,  80,  55;
  7    56, 105, 140, 154, 140,  91;
  8    84, 160, 220, 256, 260, 224, 140;
  ...
For n = 3, k = 2, a = 5, b = 12, c = 13. T(3, 2) = sqrt((13^4 - 5^4 - 12^4)/288) = 5.
		

References

  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Page 72.

Crossrefs

Cf. A120070 (b leg), A055096 (c hypotenuse).
Cf. A006414 (row sums), A000292 (column 1), A077414 (column 2), A000330 (diagonal), A107984 (transpose), A210440 (diagonal which begins with 4).

Programs

  • Mathematica
    T[n_,k_]:=n*k(n^2-k^2)/6; Table[T[n,k],{n,2,11},{k,n-1}]//Flatten (* Stefano Spezia, Jul 11 2025 *)
  • PARI
    apply( {A354968(n, k=0)=k|| k=n-1-(1-n=ceil(sqrt(8*n-7)/2+.5))*(2-n)\2; k*(n-k)*n*(n+k)\6}, [2..66]) \\ M. F. Hasler, May 08 2025

Formula

G.f.: x^2*y*(1 + x*y - 4*x^2*y + x^3*y + x^4*y^2)/((1 - x)^4*(1 - x*y)^4). - Stefano Spezia, Jul 11 2025