cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A277557 The ordered image of the 1-to-1 mapping of an integer ordered pair (x,y) into an integer using Cantor's pairing function, where 0 < x < y, gcd(x,y)=1 and x+y odd.

Original entry on oeis.org

8, 18, 19, 32, 33, 34, 50, 52, 53, 72, 73, 74, 75, 76, 98, 99, 100, 101, 102, 103, 128, 131, 133, 134, 162, 163, 164, 165, 166, 167, 168, 169, 200, 201, 202, 203, 204, 205, 206, 207, 208, 242, 244, 247, 248, 250, 251, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 338
Offset: 1

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Author

Frank M Jackson, Oct 19 2016

Keywords

Comments

The mapping of the ordered pair (x,y) to an integer uses Cantor's pairing function to generate the integer as (x+y)(x+y+1)/2+y. Also for every ordered pair (x,y) such that 0 < x < y, gcd(x,y)=1 and x+y odd, there exists a primitive Pythagorean triple (PPT) (a, b, c) such that a = y^2-x^2, b = 2xy, c = x^2+y^2. Therefore each term in the sequence represents a unique PPT.
Numbers n for which 0 < A025581(n) < A002262(n) and A025581(n)+A002262(n) is odd, and gcd(A025581(n), A002262(n)) = 1. [The definition expressed with A-numbers.] - Antti Karttunen, Nov 02 2016
See also the triangle T(y, x) with the values for PPTs given in A278147. - Wolfdieter Lang, Nov 24 2016

Examples

			a(5)=33 because the ordered pair (2,5) maps to 33 by Cantor's pairing function (see below) and is the 5th such occurrence. Also x=2, y=5 generates a PPT with sides (21,20,29).
Note: Cantor's pairing function is simply A001477 in its two-argument tabular form A001477(k, n) = n + (k+n)*(k+n+1)/2, thus A001477(2,5) = 5 + (2+5)*(2+5+1)/2 = 33. - _Antti Karttunen_, Nov 02 2016
		

Crossrefs

Cf. A020882 (is obtained when A048147(a(n)) is sorted into ascending order), A008846 (same with duplicates removed).

Programs

  • Mathematica
    Cantor[{i_, j_}] := (i+j)(i+j+1)/2+j; getparts[n_] := Reverse@Select[Reverse[IntegerPartitions[n, {2}], 2], GCD@@#==1 &]; pairs=Flatten[Table[getparts[2n+1], {n, 1, 20}], 1]; Table[Cantor[pairs[[n]]], {n, 1, Length[pairs]}]
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