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Revision History for A024406

(Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
Showing entries 1-10 | older changes

Ordered areas of primitive Pythagorean triangles.
(history; published version)

#42 by Bruno Berselli at Sat Apr 25 02:24:34 EDT 2020

STATUS

reviewed

approved

#41 by Joerg Arndt at Sat Apr 25 01:20:57 EDT 2020

STATUS

proposed

reviewed

#40 by Wolfdieter Lang at Fri Apr 24 11:49:13 EDT 2020

STATUS

editing

proposed

Discussion

Fri Apr 24

11:53

Michel Marcus: ok for me

#39 by Wolfdieter Lang at Fri Apr 24 11:49:01 EDT 2020

COMMENTS

This sequence also gives Fibonacci's congruous (or congruent) numbers (or congrua) divided by 4 with multiplicities, not regarding leg exchange in the underlying primitive Pythagorean triangle. See A258150 and the example. - Wolfdieter Lang, Jun 14 2015

LINKS

Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/CongruumProblem.html">Congruum Problem </a>

CROSSREFS

Cf. A009112, A024365, A094182, A094183, A256418 (congrua), A258150.

STATUS

proposed

editing

#38 by Michel Marcus at Thu Apr 16 11:02:39 EDT 2020

STATUS

editing

proposed

Discussion

Fri Apr 24

01:01

Joerg Arndt: IMO yes, you could ask WDL.

01:24

Michel Marcus: I asked him to come and see

06:54

Wolfdieter Lang: I suppose I used Siglers translation of Fibonacci's LQ  (BoS). He uses on p. 64  'congruous number'   for c in z^2 - y^2 = c  = y^2 - x^2 , with integers x, y , z. In  Wikipedia  https://en.wikipedia.org/wiki/Congruum this c is called 'conguum' (plural congrua). In Dickson  II , p. 459, uses 'congruent numbers' , and on p. 460 gives Leonardo da Pisa solution for c = 5 but with rational (x, y, z ) =  (31/12, 41/12, 49/12).
In my Link I used 'congruent' because it refers to  A006991: Primitive congruent numbers. Maybe we use the 'congruous number'  or 'congruous numbers' for 'congruum number'  or 'congrua numbers'  like in A256418  for c with integer (x, y, z).   In the example with  X =12*x = 31, Y = 12*y = 41 and Z =12*z = 49:   Z^2 - Y^2 = 5*12^2  = Y^2 - X^2  the congrrum number is then 5*12^2 = 720 = A256418(10). 
Conclusion: I suggest to use 'congruous  = congruum  number'  but  not equivalently to 'congruent number'.  This is different (more general), at least in MathWorld: https://mathworld.wolfram.com/CongruentNumber.html.

07:06

Michel Marcus: so you suggest to write : congruous (or congruum) number ??

11:41

Wolfdieter Lang: Yes Michel, and I add a cf. A256418 (congrua, but without multiple entries, and not only related  to primitive  PTs like here). Also I add the MathWorld link on congruum.

#37 by Michel Marcus at Thu Apr 16 11:02:29 EDT 2020

COMMENTS

This sequence also gives Fibonacci's congruous (or congruent) numbers divided by 4 with multiplicities, not regarding leg exchange in the underlying primitive Pythagorean triangle. See A258150 and the example. - Wolfdieter Lang, Jun 14 2015

STATUS

proposed

editing

Discussion

Thu Apr 16

11:02

Michel Marcus: ok like this then ?

#36 by M. F. Hasler at Thu Apr 16 10:46:44 EDT 2020

STATUS

editing

proposed

#35 by M. F. Hasler at Thu Apr 16 10:42:53 EDT 2020

FORMULA

a(n) = A121728(n)*A121729(n)/2. - M. F. Hasler, Apr 16 2020

STATUS

proposed

editing

Discussion

Thu Apr 16

10:46

M. F. Hasler: But W.Lang's LINK here also uses "congruent".... In such cases the best may be to say "... xxx (or yyy) ..." and maybe add explanation and/or link(s) to explanation(s).

#34 by Jinyuan Wang at Mon Apr 13 03:12:26 EDT 2020

STATUS

editing

proposed

Discussion

Mon Apr 13

03:16

Jinyuan Wang: ah yes, I compute hem as "congruent"...

03:29

Michel Marcus: but see A258150 comment by Wolfdieter: For the history of this problem, see Dickson, pp. 459-472 (he uses the (misleading) term congruent number).

03:32

Jinyuan Wang: but I don't see differece of congruent and congruous ? maybe after Dickson, we use congruous instead of congruent

#33 by Jinyuan Wang at Mon Apr 13 03:11:11 EDT 2020

CROSSREFS

Cf. A009112, A024365, A094182, A094183, A258150.

Discussion

Mon Apr 13

03:12

Jinyuan Wang: yes right terms, examed