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%I #67 Jun 02 2023 09:37:01
%S 1,4,5,10,16,14,20,35,40,30,35,64,81,80,55,56,105,140,154,140,91,84,
%T 160,220,256,260,224,140,120,231,324,390,420,405,336,204,165,320,455,
%U 560,625,640,595,480,285,220,429,616,770,880,935,924,836,660,385,286,560,810,1024
%N Triangle read by rows: T(n, k) = n*k*(n+k)*(n-k)/6.
%C 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.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Pythagorean_triple">Pythagorean Triple</a>.
%H <a href="/index/Ps#PyTrip">Index entries related to Pythagorean Triples</a>.
%e Triangle begins:
%e n/k 1 2 3 4 5 6 7
%e 2 1;
%e 3 4, 5;
%e 4 10, 16, 14;
%e 5 20, 35, 40, 30;
%e 6 35, 64, 81, 80, 55;
%e 7 56, 105, 140, 154, 140, 91;
%e 8 84, 160, 220, 256, 260, 224, 140;
%e ...
%e For n = 3, k = 2, a = 5, b = 12, c = 13. T(3, 2) = sqrt((13^4 - 5^4 - 12^4)/288) = 5.
%Y Cf. A120070 (b leg), A055096 (c hypotenuse).
%Y Cf. A006414 (row sums), A000292 (column 1), A077414 (column 2), A000330 (diagonal), A107984 (transpose), A210440 (diagonal which begins with 4).
%K nonn,easy,tabl
%O 2,2
%A _Ali Sada_ and _Yifan Xie_, Jun 14 2022