A081804 -id:A081804 - cp's OEIS frontend

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

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

Clark Kimberling

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

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.

David W. Wilson, Table of n, a(n) for n = 1..10000 (first 1593 terms from M. F. Hasler)
M. de Frénicle, Méthode pour trouver la solutions des problèmes par les exclusions (B.N.F. permanent link to a scan of the original edition).
Werner Hürlimann, Exact and Asymptotic Evaluation of the Number of Distinct Primitive Cuboids, Journal of Integer Sequences, Vol. 18 (2015), Article 15.2.5.
Hans Isdahl, Pythagoras site (in Norwegian). [from Internet Archive Wayback Machine]
Ron Knott, Pythagorean Triples and Online Calculators.
H. P. Robinson and N. J. A. Sloane, Correspondence, 1971-1972.
E. S. Rowland, Primitive Solutions to x^2 + y^2 = z^2.
Michael Somos, Table of primitive Pythagorean triples and related parameters.
Eric Weisstein's World of Mathematics, Pythagorean Triple.

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

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 *)

{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

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

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

A335348 Largest side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.

Original entry on oeis.org

22, 31, 41, 59, 71, 101, 109, 122, 149, 158, 178, 211, 209, 271, 262, 242, 302, 278, 341, 361, 358, 362, 419, 382, 418, 449, 478, 439, 451, 551, 482, 562, 502, 599, 541, 638, 659, 701, 638, 649, 622, 718, 781, 758, 662, 811, 698, 802, 902

Offset: 1

Bernard Schott, Jun 06 2020

nonn

If medians at A and B are perpendicular at the centroid G, then a^2 + b^2 = 5 * c^2 (see picture in Maths Challenge link), hence c is always the smallest side.
For the corresponding primitive triples and miscellaneous properties, see A335034; such a triangle with sides of integer lengths cannot be isosceles.
The largest side b with c < a < b is not divisible by 3, 4, or 5, and the odd prime factors of this largest side term b are all of the form 10*k +- 1.
In each increasing triple (c,a,b), c is the smallest odd side (A335036), but the largest side b can be either the even side (A335273) or the largest odd side (see formulas and examples for explanations).
This sequence is not increasing: a(12) = 211 for triangle with perimeter = 442 and a(13) = 209 for triangle with perimeter = 464; hence the largest side is not an increasing function of the perimeter of these triangles.

			-> For 1st class of triangles, u/v > 3:
(u,v) = (4,1), then 3 < u/v < 3+sqrt(10) and (c,a,b) = (c, a',b') = (17,22,31); the relation is 22^2 + 31^2 = 5 * 17^2 = 1445 with a(2) = 31 = b = b'
(u,v) = (10,1), then u/v > 3+sqrt(10) and (c,a,b) = (c, b' ,a') = (101, 139, 178), the relation is 139^2 + 178^2 = 5 * 101^2 = 51005 with a(11) = 178 = b = a'.
-> For 2nd class, 1 < u/v < 2:
(u,v) = (3,2), then (1+sqrt(10))/3 < u/v < 2 and (c,a,b) = (c, b', a') = (13,19,22), the relation is 19^2 + 22^2 = 5 * 13^2 = 845 with a(1) = 22 = b = a'.
(u,v) = (4,3), then 1 < u/v < (1+sqrt(10))/3 and (c,a,b) = (c, a', b') = (25,38,41); the relation is 38^2 + 41^2 = 5 * 25^2 = 3125 with a(3) = 41 = b = b'.

Maths Challenge, Perpendicular medians, Problem with picture.
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A335034 (primitive triples), A335035 (corresponding perimeters), A335036 (smallest side), A335347 (middle side), A335273 (even side).

Cf. A081804 (similar, with hypotenuse for primitive Pythagorean triples).

mycmp(x, y) = {my(xp = vecsum(x), yp = vecsum(y)); if (xp!=yp, return (xp-yp)); return (x[1] - y[1]); }
lista(nn) = {my(vm = List(), vt, w); for (u=1, nn, for (v=1, nn, if (gcd(u, v) == 1, vt = 0; if ((u/v > 3) && ((u-3*v) % 5), vt = [2*(u^2-u*v-v^2), u^2+4*u*v-v^2, u^2+v^2]); if ((u/v > 1) && (u/v < 2) && ((u-2*v) % 5), vt = [2*(u^2+u*v-v^2), -u^2+4*u*v+v^2, u^2+v^2]); if (gcd(vt) == 1, listput(vm, vt)); ); ); ); w = vecsort(apply(vecsort, Vec(vm)); , mycmp); vector(#w, k, w[k][3]); }  \\ Michel Marcus, Jun 06 2020

a(n) = A335034(3n).
a(n) = A335035(n) - A335036(n) - A335347(n).
There exist two disjoint classes of such triangles, obtained with two distinct families of formulas: let u > v > 0 , u and v with different parities, gcd(u,v) = 1; a' is the even side and b' the largest odd side.
--> 1st class of triangles: (a',b',c) = (2*(u^2-uv-v^2), u^2+4*u*v-v^2, u^2+v^2) with u/v > 3 and 5 doesn't divide u-3v.
If 3 < u/v < 3+sqrt(10) then a' (even) < b' and the triple in increasing order is (c, a = a', b = b'),
if u/v > 3+sqrt(10) then a' (even) > b' and the triple in increasing order is (c, a = b', b = a').
--> 2nd class of triangles: (a',b',c) = (2*(u^2+uv-v^2), -u^2+4*u*v+v^2, u^2+v^2) with 1 < u/v < 2 and 5 doesn't divide u-2v.
If 1 < u/v < (1+sqrt(10))/3 then a' (even) < b' and the triple in increasing order is (c, a = a', b = b'),
If (1+sqrt(10))/3 < u/v < 2 then a' (even) > b' and the triple in increasing order is (c, a = b', b = a').

A349536 Consider a circle on the Z X Z lattice with radius equal to the Pythagorean hypotenuse h(n) (A009003); a(n) = number of Pythagorean triples inside a Pi/4 sector of the circle.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26, 27, 31, 32, 33, 34, 35, 37, 38, 39, 40, 44, 45, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 71, 75, 76, 77, 78, 79, 80, 84, 85, 86, 87, 89

Offset: 1

Alexander Kritov, Nov 21 2021

nonn

Number of Pythagorean triples with hypotenuse less than or equal to the next one.

			The count of non-primitive Pythagorean triples as they appear in order of increasing hypotenuse:
.
       Hypotenuse
   n  (A009003(n))       Sides       a(n)
  --  ------------  ---------------  ----
   1        5            (3,4)         1
   2       10            (6,8)         2
   3       13            (5,12)        3
   4       15            (9,12)        4
   5       17            (8,15)        5
   6       20           (12,16)        6
   7       25       (7,24), (15,20)    8
   8       26           (10,24)        9
   9       29           (20,21)       10

W. Sierpinski, Pythagorean Triangles, Dover Publications, 2003.

Alexander Kritov, Table of n, a(n) for n = 1..1050
Manuel Benito and Juan L. Varona, Pythagorean triangles with legs less than n, Journal of Computational and Applied Mathematics 143, (2002), pp. 117-126.
E. Eckert, The group of primitive Pythagorean triangles, Mathematics Magazine 57 (1984) 22-27.
Alexander Kritov, C code that generates b-file
Eric Weisstein's World of Mathematics, Pythagorean Triple

Cf. A008846, A009003.
Cf. A020882, A081804, A008846, A020883, A046086.
Cf. A349538 (extension to the full circle of Z^2 lattice).

// see enclosed main.c
for (long j=1;j< 101;++j)
{
for (long k=1;k< 101;++k)
{
if (k<=j)   // to avoid pairs (as we need 1/8 or quarter plane)
    {
          double hyp=sqrt(j*j+k*k);
          double c= (double) floor (hyp );
if   (fabs(hyp - c) < DBL_EPSILON)  arr[r++]= (long) c;
}}}
bubbleSort(arr, r);//sort by hypotenuse increase
for (long j=0;j< r;++j)
{
   if  ( arr[j] != arr[j+1] )
    {
        // write to file: j is the sequence value a[n]*2
        // arr[j] is the hypotenuse value
    }
}

Conjecture: the increment is a(n+1) - a(n) = 2^(m-1), where m is the sum of all powers of the Pythagorean primes (A002144) in the factorization of hypotenuse h(n+1) (see Eckert for PPT). However, starting from 58 the increment is 3.

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

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A335348 Largest side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.

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A349536 Consider a circle on the Z X Z lattice with radius equal to the Pythagorean hypotenuse h(n) (A009003); a(n) = number of Pythagorean triples inside a Pi/4 sector of the circle.

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