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Revision History for A358210

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Showing entries 1-10 | older changes

Congruent number sequence starting from the Pythagorean triple (3,4,5).
(history; published version)

#16 by N. J. A. Sloane at Wed Dec 21 21:59:03 EST 2022

STATUS

proposed

approved

#15 by Michel Marcus at Fri Nov 04 14:30:44 EDT 2022

STATUS

editing

proposed

Discussion

Sat Dec 17

14:02

Sean A. Irvine: This is not clear enough to me. What recurrence is the example referring to? In particular, how do you go from (3,4,5) to (15/2,4,17/2)?

Mon Dec 19

19:33

Gerry Martens: OK : Starting from 
{a, b, c} = {3, 4, 5}
{n, p, q} = {a b /2, a, 1} = {6,3,1}
{n1, p1, q1} = {Sqrt[p^4 + 4 n^2 q^4], p Sqrt[p^4 + 4 n^2 q^4], q^2 n} = {15,45,6}
{a1, b1, c1} = {p1/q1, 2 n1 q1/p1, Sqrt[p1^4 + 4 n1^2 q1^4]/(p1 q1)} = {15/2,4,17/2}

Repeating the recurrence for the b-side :
{n, p, q} = {a1 b1 /2, Numerator[b1], Denominator[b1]} = {15,4,1}
{n1, p1, q1} = {Sqrt[p^4 + 4 n^2 q^4], p Sqrt[p^4 + 4 n^2 q^4], q^2 n} ={34,136,15}
{a1, b1, c1} = {p1/q1, 2 n1 q1/p1, Sqrt[p1^4 + 4 n1^2 q1^4]/(p1 q1)}= {136/15,15/2,353/30}
....
 I hope this is making it more clear to you.

19:37

Gerry Martens: Omg the spacing looks terrible , it will be more readable inserting a linefeed character before each assignments {..}=

Tue Dec 20

12:08

N. J. A. Sloane: Could you please add the recurrence, perhaps as a comment?

Wed Dec 21

21:59

N. J. A. Sloane: I don't want to hold this up any longer, but it would be nice if someone could add the recurrence,

#14 by Michel Marcus at Fri Nov 04 14:30:40 EDT 2022

CROSSREFS

Cf. A081465 (numerators of hypotenuses offset 1).

STATUS

proposed

editing

#13 by Gerry Martens at Fri Nov 04 13:48:04 EDT 2022

STATUS

editing

proposed

Discussion

Fri Nov 04

14:00

Michel Marcus: offset 1 in xref can go ??

14:20

Gerry Martens: In case you take into account the cf. sequence offset then yes

#12 by Gerry Martens at Fri Nov 04 13:47:39 EDT 2022

EXAMPLE

Starting with the triple (3,4,5) and choosing the b side we obtain by the recurrence choosing the b side the right triangles: (15/2, 4, 17/2), (136/15, 15/2, 353/30), (5295/136, 272/15, 87617/2040), (47663648/79425, 79425/136, 9045146753/10801800), ...

Discussion

Fri Nov 04

13:48

Gerry Martens: How is this?

#11 by Gerry Martens at Fri Nov 04 13:40:25 EDT 2022

EXAMPLE

So a(34) = (5295/136) * (272/15) / 2 = 353.

STATUS

proposed

editing

#10 by Gerry Martens at Fri Nov 04 13:19:36 EDT 2022

STATUS

editing

proposed

Discussion

Fri Nov 04

13:28

Michel Marcus: So a(3)  : should be changed

13:28

Michel Marcus: we obtain by the recurrence choosing the b side the right triangles : a word is missing ?

#9 by Gerry Martens at Fri Nov 04 13:02:58 EDT 2022

OFFSET

0,1,1

STATUS

proposed

editing

Discussion

Fri Nov 04

13:19

Gerry Martens: Yes I think you are right looking at A081465

#8 by Michel Marcus at Fri Nov 04 11:21:53 EDT 2022

STATUS

editing

proposed

#7 by Michel Marcus at Fri Nov 04 11:21:03 EDT 2022

EXAMPLE

Starting with triple (3,4,5) we obtain by the recurrence choosing the b side the right triangles: (15/2, 4, 17/2), (136/15, 15/2, 353/30), (5295/136, 272/15, 87617/2040), (47663648/79425, 79425/136, 9045146753/10801800), ...

(15/2, 4, 17/2), (136/15, 15/2, 353/30), (5295/136, 272/15, 87617/2040), (47663648/79425, 79425/136, 9045146753/10801800), ...

Discussion

Fri Nov 04

11:21

Michel Marcus: I wonder if offset should rather be 1 ?  since it is a list ??