Revision History for A343769
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A343769
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Sorted areas of primitive Heronian triangles for which a rectangle exists with integer dimensions and with perimeter and area equal respectively to the perimeter and area of the triangle.
(history;
published version)
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#8 by N. J. A. Sloane at Sat Oct 09 07:41:54 EDT 2021
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#7 by Michel Marcus at Thu Sep 09 13:47:14 EDT 2021
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#6 by Michel Marcus at Thu Sep 09 13:47:08 EDT 2021
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| EXAMPLE
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a(1) = 12 because 12 is the area of the 5-5-6 triangle, which is the least-area primitive Heronian triangle for which a rectangle exists with integer dimensions (2-by-6) and with perimeter (16) and area (12) equal respectively to the perimeter and area of the triangle. a(2) = 126 because 126 is the area of the 13-20-21 triangle, which is the second-least-area primitive Heronian triangle for which a rectangle exists with integer dimensions (6-by-21) and with perimeter (54) and area (126) equal respectively to the perimeter and area of the triangle.
a(2) = 126 because 126 is the area of the 13-20-21 triangle, which is the second-least-area primitive Heronian triangle for which a rectangle exists with integer dimensions (6-by-21) and with perimeter (54) and area (126) equal respectively to the perimeter and area of the triangle.
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| KEYWORD
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nonn,more,changed
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| STATUS
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proposed
editing
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#5 by Jason Zimba at Thu Sep 09 12:22:58 EDT 2021
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Discussion
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Thu Sep 09
| 13:45
| Michel Marcus: Well I was just wondering
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#4 by Jason Zimba at Wed Sep 08 08:50:32 EDT 2021
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| EXAMPLE
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a(1) = 12 because 12 is the area of the 5-5-6 triangle, which is the least-area primitive Heronian triangle for which a rectangle exists with integer dimensions (2-by-6) and with perimeter (16) and area (12) equal respectively to the perimeter and area of the triangle. a(2) = 126 because 126 is the area of the 13-20-21 triangle, which is the second-least-area primitive Heronian triangle for which a rectangle exists with integer dimensions (6-by-21) and with perimeter (54) and area (126) equal respectively to the perimeter and area of the triangle.
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| STATUS
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proposed
editing
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#3 by Jason Zimba at Wed Sep 08 08:48:52 EDT 2021
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#2 by Jason Zimba at Wed Apr 28 17:56:39 EDT 2021
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| NAME
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allocatedSorted areas of primitive Heronian triangles for which a rectangle exists with integer dimensions and with perimeter and area equal respectively to the perimeter and area of Jasonthe Zimbatriangle.
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| DATA
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12, 126, 624, 1260, 1800, 2100, 2850, 4536, 5292, 5580, 8820, 9900, 12600, 12642, 14850, 15600, 17640, 19110, 21756, 23400, 24948, 25200, 25536, 28350, 47040, 47304
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| OFFSET
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1,1
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| LINKS
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Jason Zimba, <a href="https://jzimba.blogspot.com/2021/04/there-are-infinitely-many-rectangular.html">There are infinitely many rectangular Heronian triangles</a>.
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| EXAMPLE
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a(1) = 12 because 12 is the area of the 5-5-6 triangle, which is the least-area primitive Heronian triangle for which a rectangle exists with integer dimensions (2-by-6) and with perimeter (16) and area (12) equal respectively to the perimeter and area of the triangle. a(2) = 126 because 126 is the area of the 13-20-21 triangle, which is the least-area primitive Heronian triangle for which a rectangle exists with integer dimensions (6-by-21) and with perimeter (54) and area (126) equal respectively to the perimeter and area of the triangle.
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| MATHEMATICA
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(* Adapted from Albert Lau's program for A224301 *)
AMax = 10000;
Do[c = p/b;
a1 = Sqrt[b^2 + c^2 + 2 Sqrt[b^2 c^2 - 4 A^2]];
a2 = Sqrt[b^2 + c^2 - 2 Sqrt[b^2 c^2 - 4 A^2]];
If[IntegerQ[a2] && GCD[a2, b, c] == 1 &&
a1 > a2 >= b && (per = a2 + b + c;
IntegerQ[(per + Sqrt[per^2 - 16 A])/4]), A // Sow(*{A, a2, b, c}//
Sow*)];
If[IntegerQ[a1] &&
GCD[a1, b, c] == 1 && (per = a1 + b + c;
IntegerQ[(per + Sqrt[per^2 - 16 A])/4]), A // Sow(*{A, a1, b, c}//
Sow*)]; , {A, 6, AMax, 6}, {p,
4 A^2 // Divisors //
Select[#, EvenQ[#] && # >= 2 A &] & // #/2 + 2 A^2/# & //
Select[#, IntegerQ] &}, {b,
p // Divisors // Select[#, #^2 >= p &] &}] // Reap // Last // Last
{a1, a2, c} =.;
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| CROSSREFS
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Subsequence of A224301.
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| KEYWORD
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allocated
nonn
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| AUTHOR
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Jason Zimba, Apr 28 2021
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| STATUS
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approved
editing
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Discussion
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Wed May 05
| 18:38
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Wed May 12
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Wed Sep 08
| 08:33
| Michel Marcus: why don't you submit your sequence ???
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| 08:48
| Jason Zimba: Well I drafted the sequence and was going to submit it for review, but then I thought I should wait until I had obtained some form of peer review first. I have a problem upcoming in the College Math Journal that I thought would be a good citation. In the meantime, I get a weekly reminder from the OEIS server, which I simply ignore. But I apologize. I had assumed that this process was confined only to me plus 1 robot...did not realize that editors would be having to deal with it. I'll go ahead and submit it. Best, Jason
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#1 by Jason Zimba at Wed Apr 28 17:56:39 EDT 2021
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| NAME
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allocated for Jason Zimba
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| KEYWORD
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allocated
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| STATUS
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approved
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