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.

A020882 Ordered hypotenuses (with multiplicity) of primitive Pythagorean triangles.

Original entry on oeis.org

5, 13, 17, 25, 29, 37, 41, 53, 61, 65, 65, 73, 85, 85, 89, 97, 101, 109, 113, 125, 137, 145, 145, 149, 157, 169, 173, 181, 185, 185, 193, 197, 205, 205, 221, 221, 229, 233, 241, 257, 265, 265, 269, 277, 281, 289, 293, 305, 305, 313, 317, 325, 325, 337, 349, 353, 365, 365
Offset: 1

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Comments

The largest member 'c' of the primitive Pythagorean triples (a,b,c) ordered by increasing c.
These are numbers of the form a^2 + b^2 where gcd(b-a, 2*a*b)=1. - M. F. Hasler, Apr 04 2010
Equivalently, numbers of the form a^2 + b^2 where gcd(a,b) = 1 and a and b are not both odd. To avoid double-counting, require a > b > 0. - Franklin T. Adams-Watters, Mar 15 2015
The density of such points in a circle with radius squared = a(n) is ~ Pi * a(n). Restricting to a > b > 0 reduces this by a factor of 1/8; requiring gcd(a,b)=1 provides a factor of 6/Pi^2; and a, b not both odd is a factor of 2/3. (2/3, not 3/4, because the case a, b both even has already been eliminated.) Multiplying, a(n) * Pi * 1/8 * 6/Pi^2 * 2/3 is a(n) / (2 * Pi). But n is approximately this number of points, so a(n) ~ 2 * Pi * n. Conjectured by David W. Wilson, proof by Franklin T. Adams-Watters, Mar 15 2015
Permutations are in A094194, A088511, A121727, A119321, A113482 and A081804. Entries of A024409 occur here more than once. - R. J. Mathar, Apr 12 2010
The distinct terms of this sequence seem to constitute a subset of the sequence defined as a(n) = (-1)^n + 6*n for n >= 1. - Alexander R. Povolotsky, Mar 15 2015
The terms in this sequence are given by f(m,n) = m^2 + n^2 where m and n are any two integers satisfying m > 1, n < m, the greatest common divisor of m and n is 1, and m and n are both not odd. E.g., f(m,n) = f(2,1) = 2^2 + 1^2 = 4 + 1 = 5. - Agola Kisira Odero, Apr 29 2016

References

  • M. de Frénicle, "Méthode pour trouver la solutions des problèmes par les exclusions", in: "Divers ouvrages de mathématiques et de physique, par Messieurs de l'Académie royale des sciences", Paris, 1693, pp 1-44.

Crossrefs

Cf. A004613, A008846, A020883-A020886, A046086, A046087, A222946 (as a number triangle).

Programs

  • Mathematica
    t={};Do[Do[a=Sqrt[c^2-b^2];If[a>b,Break[]];If[IntegerQ[a]&&GCD[a,b,c]==1,AppendTo[t,c]],{b,c-1,3,-1}],{c,400}];t (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
    f[c_] := Block[{a = 1, b, lst = {}}, While[b = Sqrt[c^2 - a^2]; a < b, If[ IntegerQ@ b && GCD[a, b, c] == 1, AppendTo[lst, a]]; a++]; lst]
    Join @@ Table[ConstantArray[n, Length@f@n], {n, 1, 400, 4}] (* Robert G. Wilson v, Mar 16 2014; corrected by Andrey Zabolotskiy, Oct 31 2019 *)
  • PARI
    {my( c=0, new=[]); for( b=1,99, for( a=1, b-1, gcd(b-a,2*a*b) == 1 && new=concat(new,a^2+b^2)); new=vecsort(new); for( j=1,#new, new[j] > (b+1)^2 & (new=vecextract(new, Str(j,".."))) & next(2); write("b020882.txt",c++," "new[j])); new=[])} \\ M. F. Hasler, Apr 04 2010

Formula

a(n) = sqrt((A120681(n)^2 + A120682(n)^2)/2). - Lekraj Beedassy, Jun 24 2006
a(n) = sqrt(A046086(n)^2 + A046087(n)^2). - Zak Seidov, Apr 12 2011
a(n) ~ 2*Pi*n. - observation by David W. Wilson, proved by Franklin T. Adams-Watters (cf. comments), Mar 15 2015
a(n) = sqrt(A180620(n)^2 + A231100(n)^2). - Rui Lin, Oct 09 2019

Extensions

Edited by N. J. A. Sloane, May 15 2010