[seqfan] Re: A sequence connected to the Pythagorean triples
Yifan Xie
xieyifan4013 at 163.com
Mon Jun 13 04:44:45 CEST 2022
Hi,
I think it's better to write the terms in a triangle T(m, k) satisfying m>0, k>0, k<m.
The triangle begins:
1,
4, 5
10, 16, 14
20, 35, 40, 30
35, 64, 81, 80, 55
...
Is this clearer?
I also checked the row sums, which begins in
1, 9, 40, 125, 315,
is A006414.
I couldn't find the term 9.
Best regards,
Yifan Xie
At 2022-06-12 15:06:12, "Ali Sada via SeqFan" <seqfan at list.seqfan.eu> wrote:
>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>
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