A383615
Length of the long leg of the unique primitive Pythagorean triple (x,y,z) such that (x-y+z)/2 = A000108(n) and its long leg and hypotenuse are consecutive natural numbers.
Original entry on oeis.org
4, 40, 364, 3444, 34584, 367224, 4086940, 47268364, 564177640, 6911470020, 86537568264, 1103799334200, 14305253278320, 187980019758360, 2500329584942460, 33615542888998620, 456277454520102600, 6246438361923425820, 86175353763393711960, 1197196443738946826760
Offset: 2
Triangles begin:
n=2: 3, 4, 5;
n=3: 9, 40, 41.
This sequence is column 2.
A383616
Semiperimeter of the unique primitive Pythagorean triple (a,b,c) such that (a-b+c)/2 is A000108(n) and such that its long leg and its hypotenuse are consecutive natural numbers.
Original entry on oeis.org
1, 1, 6, 45, 378, 3486, 34716, 367653, 4088370, 47273226, 564194436, 6911528806, 86537776276, 1103800077100, 14305255952760, 187980029453205, 2500329620300130, 33615543018643410, 456277454997741300, 6246438363690689010, 86175353769957832380, 1197196443763413093780, 16738118900201817535560
Offset: 0
For n=3, the short leg is A383615(3,1) = 3, the long leg is A383615(3,2) = 4 and the hypotenuse is A383615(3,3) = 5 so the semiperimeter is then a(3) = (3 + 4 + 5)/2 = 6.
- Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2025.
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a=Table[(2n)!/(n!(n+1)!),{n,0,22}];Apply[Join,Map[{#(2#-1)}&,a]]
A383958
Sum of the legs of the unique primitive Pythagorean triple (a,b,c) such that (a-b+c)/2 = A000108(n) and its long leg and hypotenuse are consecutive natural numbers.
Original entry on oeis.org
1, 1, 7, 49, 391, 3527, 34847, 368081, 4089799, 47278087, 564211231, 6911587591, 86537984287, 1103800819999, 14305258627199, 187980039148049, 2500329655657799, 33615543148288199, 456277455475379999, 6246438365457952199, 86175353776521952799, 1197196443787879360799, 16738118900293300099199
Offset: 0
For n=3, the short leg is A383615(3,1) = 3 and the long leg is A383615(3,2) = 4 so the sum of the legs is then a(3) = 3 + 4 = 7.
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a=Table[(2n)!/(n!(n+1)!),{n,0,23}];Apply[Join,Map[{2#^2-1}&,a]]
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