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

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

2,2

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.

References

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

Links

M. F. Hasler, Table of n, a(n) for n = 2..1000, May 08 2025

Wikipedia, Pythagorean Triple.

Index entries related to Pythagorean Triples.

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

Example

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.

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 *)

Prog

(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

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).

Sequence in context: A287578 A135104 A131780 * A102006 A118735 A002970

Adjacent sequences: A354965 A354966 A354967 * A354969 A354970 A354971

Keyword

nonn,easy,tabl

Author

Ali Sada and Yifan Xie, Jun 14 2022

Status

approved