# [seqfan] A sequence connected to the Pythagorean triples

Sat Jun 11 04:03:46 CEST 2022

```Hi everyone,

Just wanted to check if this sequence is suitable for the OEIS.

a(n) = ((c^4-a^4-b^4)/288)^1/2, where a, b, and c are the Pythagorean triples ordered by increasing c.

Let S be c^4-a^4-b^4
Using the Pythagorean triples generators
a = 2mk
b = m^2-k^2
c = m^2+k^2
Then we substitute to get the value of S in terms of m and k
S = 8*m^2*k^2*((m^2-k^2))^2
That’s a multiple of 288. When we divide by 288, we get

1, 16, 25, 81, 100, 256, 625, 196, 400, 1225, 1296, 1600, 2401, 1225, 2025, 4096, 900, 6561, 10000, 3136, 8100, 6400, 11025, 14641, 19600, 20736, 3025, 28561, 23716, 15625, 7056, 25600, 38416, 48400, 19600, 50625, 15876, 32400, 65536, 14400, 83521, 62500, 53361, 8281, 99225, 67600, 104976, 60025, 130321, 152100, 160000, 50176, 27225, 129600, 102400, 194481, 176400, 207025, 234256, 99225, 19600, 279841, 313600, 164025, 240100, 331776

We take the square roots, and we get a(n)

1, 4, 5, 9, 10, 16, 25, 14, 20, 35, 36, 40, 49, 35, 45, 64, 30, 81, 100, 56, 90, 80, 105, 121, 140, 144, 55, 169, 154, 125, 84, 160, 196, 220, 140, 225, 126, 180, 256, 120, 289, 250, 231, 91, 315, 260, 324, 245, 361, 390, 400, 224, 165, 360, 320, 441, 420, 455, 484, 315, 140, 529, 560, 405, 490, 576
Best,
Ali

```