login
Triangle read by rows of Cantor pairing function value determining primitive Pythagorean triangles or 0 if there is no such triangle.
1

%I #12 Nov 23 2016 12:56:02

%S 8,0,18,19,0,32,0,33,0,50,34,0,0,0,72,0,52,0,73,0,98,53,0,74,0,99,0,

%T 128,0,75,0,100,0,0,0,162,76,0,101,0,0,0,163,0,200,0,102,0,131,0,164,

%U 0,201,0,242,103,0,0,0,165,0,202,0,0,0,288,0,133,0,166,0,203,0,244,0,289,0,338,134,0,167,0,204

%N Triangle read by rows of Cantor pairing function value determining primitive Pythagorean triangles or 0 if there is no such triangle.

%C This entry is inspired by the increasingly ordered nonvanishing entries given in A277557.

%C A primitive Pythagorean triangle is characterized by the pair [n,m], 1 <= m < n, GCD(n,m) = 1 and n+m is odd. The present triangle gives the values T(n, m) = Cantor(m,n) where Cantor(x,y) = (x+y)*(x+y+1)/2 + y. See A277557, also for links.

%C Because the Cantor pairing function N x N -> N is bijective (N = positive integers), all nonzero entries of this triangle appear only once, but here not all positive integers appear.

%C Note that in this triangle in each row the nonvanishing entries increase, but in the first rows up to some n not all T(n, m) values smaller than T(n,n-1) are covered.

%C For the area values of primitive Pythagorean triangles see the table A249869 also for comments on these triangles and references.

%F T(n, m) = (m+n)*(m+n+1)/2 + n, n >= 2, m = 1, 2, ..., n-1, and 0 if GCD(n,m) > 1 or n+m is even.

%e The triangle begins:

%e n\m 1 2 3 4 5 6 7 8 9 10...

%e 2: 8

%e 3: 0 18

%e 4: 19 0 32

%e 5: 0 33 0 50

%e 6: 34 0 0 0 7272

%e 7: 0 52 0 73 0 98

%e 8: 53 0 74 0 99 0 128

%e 9: 0 75 0 100 0 0 0 162

%e 10: 76 0 101 0 0 0 163 0 200

%e 11: 0 102 0 131 0 164 0 201 0 242

%e ...

%e n = 12: 103 0 0 0 165 0 202 0 0 0 288,

%e n = 13: 0 133 0 166 0 203 0 244 0 289 0 338,

%e n = 14: 134 0 167 0 204 0 0 0 290 0 339 0 392,

%e n = 15: 0 168 0 205 0 0 0 291 0 0 0 0 0 450.

%e ...

%e T(3,1) = 0 because 3+1 =4 is even.

%e T(4,2) = 0 because GCD(4,2) = 2 > 1.

%e T(3,2) = (2+3)*(2+3)/2 + 3 = 5*3 + 3 = 18.

%e ...

%e In order to reach all values T(n,m) <= 50 one has to take rows n = 2..6.

%e ...

%Y Cf. A277557, A249869.

%K nonn,tabl,easy

%O 2,1

%A _Wolfdieter Lang_, Nov 21 2016