A084645 -id:A084645 - cp's OEIS frontend

A009003 Hypotenuse numbers (squares are sums of 2 nonzero squares).

5, 10, 13, 15, 17, 20, 25, 26, 29, 30, 34, 35, 37, 39, 40, 41, 45, 50, 51, 52, 53, 55, 58, 60, 61, 65, 68, 70, 73, 74, 75, 78, 80, 82, 85, 87, 89, 90, 91, 95, 97, 100, 101, 102, 104, 105, 106, 109, 110, 111, 113, 115, 116, 117, 119, 120, 122, 123, 125, 130, 135, 136, 137, 140

Offset: 1

David W. Wilson

nonn

Multiples of Pythagorean primes A002144 or of primitive Pythagorean triangles' hypotenuses A008846. - Lekraj Beedassy, Nov 12 2003
This is exactly the sequence of positive integers with at least one prime divisor of the form 4k + 1. Compare A072592. - John W. Layman, Mar 12 2008 and Franklin T. Adams-Watters, Apr 26 2009
Circumradius R of the triangles such that the area, the sides and R are integers. - Michel Lagneau, Mar 03 2012
The 2 squares summing to a(n)^2 cannot be equal because sqrt(2) is not rational. - Jean-Christophe Hervé, Nov 10 2013
Closed under multiplication. The primitive elements are those with exactly one prime divisor of the form 4k + 1 with multiplicity one, which are also those for which there exists a unique integer triangle = A084645. - Jean-Christophe Hervé, Nov 11 2013
a(n) are numbers whose square is the mean of two distinct nonzero squares. This creates 1-to-1 mapping between a Pythagorean triple and a "Mean" triple. If the Pythagorean triple is written, abnormally, as {j, k, h} where j^2 +(j+k)^2 = h^2, and h = a(n), then the corresponding "Mean" triple with the same h is {k, 2j, h} where (k^2 + (k+2j)^2)/2 = h^2. For example for h = 5, the Pythagorean triple is {3, 1, 5} and the Mean triple is {1, 6, 5}. - Richard R. Forberg, Mar 01 2015
Integral side lengths of rhombuses with integral diagonals p and q (therefore also with integral areas A because A = pq/2 is some multiple of 24). No such rhombuses are squares. - Rick L. Shepherd, Apr 09 2017
Conjecture: these are bases n in which exists an n-adic integer x satisfying x^5 = x, and 5 is the smallest k>1 such that x^k =x (so x^2, x^3 and x^4 are not x). Example: the 10-adic integer x = ...499879186432 (A120817) satisfies x^5 = x, and x^2, x^3, and x^4 are not x, so 10 is in this sequence. See also A120817, A210850 and A331548. - Patrick A. Thomas, Mar 01 2020
Didactic comment: When students solve a quadratic equation a*x^2 + b*x + c = 0 (a, b, c: integers) with the solution formula, they often make the mistake of calculating b^2 + 4*a*c instead of b^2 - 4*a*c (especially if a or c is negative). If the root then turns out to be an integer, they feel safe. This sequence lists the absolute values of b for which this error can happen. Reasoning: With p^2 = b^2 - 4*a*c and q^2 = b^2 + 4*a*c it follows by addition immediately that p^2 + q^2 = 2*b^2. If 4*a*c < 0, let p = x + y and q = x - y. If 4*a*c > 0, let p = x - y and q = x + y. In both cases follows that y^2 + x^2 = b^2. So every Pythagorean triple gives an absolute value of b for which this error can occur. Example: From (y, x, b) = (3, 4, 5) follows (q^2, b^2, p^2) = (1, 25, 49) or (p^2, b^2, q^2) = (1, 25, 49) with abs(4*a*c) = 24. - Felix Huber, Jul 22 2023
Conjecture: Numbers m such that the limit: Limit_{s->1} zeta(s)*Sum_{k=1..m} [k|m]*A008683(k)*(i^k)/(k^(s - 1)) exists, which is equivalent to numbers m such that abs(Sum_{k=1..m} [k|m]*A008683(k)*(i^k)) = 0. - Mats Granvik, Jul 06 2024

Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 98-104.

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
R. Chapman, Pythagorean triples and sums of squares
Steven R. Finch, Landau-Ramanujan Constant [Broken link]
Steven R. Finch, Landau-Ramanujan Constant [From the Wayback machine]
Ron Knott, Pythagorean Triples and Online Calculators
J. Pahikkala, On contraharmonic mean and Pythagorean triples, Elemente der Mathematik, 65:2 (2010), 62-67.
Patrick A. Thomas, Solutions of x^5=x up to base 100
Index entries for sequences related to sums of squares

Cf. A000404 (sums of 2 squares), A004431 (sums of 2 distinct squares), A009000 (hypotenuse numbers with repetition), A072592, A004613, A187811.
Complement of A004144. Primes in this sequence give A002144. Same as A146984 (integer contraharmonic means) as sets - see Pahikkala 2010, Theorem 5.
Cf. A083025, A084645 (primitive elements), A084646, A084647, A084648, A084649, A006339.

import Data.List (findIndices)
a009003 n = a009003_list !! (n-1)
a009003_list = map (+ 1) $ findIndices (> 0) a005089_list
-- Reinhard Zumkeller, Jan 07 2013

isA009003 := proc(n)
    local p;
    for p in numtheory[factorset](n) do
        if modp(p,4) = 1 then
            return true;
        end if;
    end do:
    false;
end proc:
for n from 1 to 200 do
    if isA009003(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Nov 17 2014

f[n_] := Module[{k = 1}, While[(n - k^2)^(1/2) != IntegerPart[(n - k^2)^(1/2)], k++; If[2 * k^2 >= n, k = 0; Break[]]]; k]; A009003 = {}; Do[If[f[n^2] > 0, AppendTo[A009003, n]], {n, 3, 100}]; A009003 (* Vladimir Joseph Stephan Orlovsky, Jun 15 2009 *)
Select[Range[200], Length[PowersRepresentations[#^2, 2, 2]] > 1 &] (* Alonso del Arte, Feb 11 2014 *)

is_A009003(n)=setsearch(Set(factor(n)[,1]%4),1)  \\ M. F. Hasler, May 27 2012

list(lim)=my(v=List(),u=vectorsmall(lim\=1)); forprimestep(p=5,lim,4, forstep(n=p,lim,p, u[n]=1)); for(i=5,lim, if(u[i], listput(v,i))); u=0; Vec(v) \\ Charles R Greathouse IV, Jan 13 2022

from itertools import count, islice
from sympy import primefactors
def A009003_gen(): # generator of terms
    return filter(lambda n:any(map(lambda p: p % 4 == 1,primefactors(n))),count(1))
A009003_list = list(islice(A009003_gen(),20)) # Chai Wah Wu, Jun 22 2022

A005089(a(n)) > 0. - Reinhard Zumkeller, Jan 07 2013
a(n) ~ n. - Charles R Greathouse IV, Jan 13 2022
a(n) = sqrt(n-th square in A000404), where A000404 lists the sums of two nonzero squares. - M. F. Hasler, Jun 20 2025

Definition edited by Jean-Christophe Hervé, Nov 10 2013

A046080 a(n) is the number of integer-sided right triangles with hypotenuse n.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 2, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 2, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 4, 0, 0, 1, 0, 1, 0, 0, 1, 1, 2, 0, 0, 1, 0, 1, 0, 1, 0, 0, 4, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0

Offset: 1

Eric W. Weisstein

nonn

Or number of ways n^2 can be written as the sum of two positive squares: a(5) = 1: 3^2 + 4^2 = 5^2; a(25) = 2: 7^2 + 24^2 = 15^2 + 20^2 = 25^2. - Alois P. Heinz, Aug 01 2019

A. H. Beiler, Recreations in the Theory of Numbers, New York: Dover, pp. 116-117, 1966.

Stanislav Sykora, Table of n, a(n) for n = 1..20000
Ron Knott, Pythagorean Triples and Online Calculators
F. Richman, Pythagorean Triples
A. Tripathi, On Pythagorean triples containing a fixed integer, Fib. Q., 46/47 (2008/2009), 331-340. See Theorem 7.
Eric Weisstein's World of Mathematics, Pythagorean Triple

First differs from A083025 at n=65.
Cf. A000290, A006339, A024362, A046079, A046081, A009000.
A088111 gives records; A088959 gives where records occur.
Cf. A046109, A063725.
Cf. A004144, A084647, A084648, A084649.
Partial sums: A224921.

f:= proc(n) local F,t;
  F:= select(t -> t[1] mod 4 = 1, ifactors(n)[2]);
  1/2*(mul(2*t[2]+1, t=F)-1)
end proc:
map(f, [$1..100]); # Robert Israel, Jul 18 2016

a[1] = 0; a[n_] := With[{fi = Select[ FactorInteger[n], Mod[#[[1]], 4] == 1 & ][[All, 2]]}, (Times @@ (2*fi+1)-1)/2]; Table[a[n], {n, 1, 99}] (* Jean-François Alcover, Feb 06 2012, after first formula *)

a(n)={my(m=0,k=n,n2=n*n,k2,l2);
while(1,k=k-1;k2=k*k;l2=n2-k2;if(l2>k2,break);if(issquare(l2),m++));return(m)} \\ brute force, Stanislav Sykora, Mar 18 2015

{a(n) = if( n<1, 0, sum(k=1, sqrtint(n^2 \ 2), issquare(n^2 - k^2)))}; /* Michael Somos, Mar 29 2015 */

a(n) = {my(f = factor(n/(2^valuation(n, 2)))); (prod(k=1, #f~, if ((f[k,1] % 4) == 1, 2*f[k,2] + 1, 1)) - 1)/2;} \\ Michel Marcus, Mar 08 2016

from math import prod
from sympy import factorint
def A046080(n): return prod((e<<1)+1 for p,e in factorint(n).items() if p&3==1)>>1 # Chai Wah Wu, Sep 06 2022

Let n = 2^e_2 * product_i p_i^f_i * product_j q_j^g_j where p_i == 1 mod 4, q_j == 3 mod 4; then a(n) = (1/2)*(product_i (2*f_i + 1) - 1). - Beiler, corrected
8*a(n) + 4 = A046109(n) for n > 0. - Ralf Stephan, Mar 14 2004
a(n) = 0 for n in A004144. - Lekraj Beedassy, May 14 2004
a(A084645(k)) = 1. - Ruediger Jehn, Jan 14 2022
a(A084646(k)) = 2. - Ruediger Jehn, Jan 14 2022
a(A084647(k)) = 3. - Jean-Christophe Hervé, Dec 01 2013
a(A084648(k)) = 4. - Jean-Christophe Hervé, Dec 01 2013
a(A084649(k)) = 5. - Jean-Christophe Hervé, Dec 01 2013
a(n) = A063725(n^2) / 2. - Michael Somos, Mar 29 2015
a(n) = Sum_{k=1..n} Sum_{i=1..k} [i^2 + k^2 = n^2], where [ ] is the Iverson bracket. - Wesley Ivan Hurt, Dec 10 2021
a(A002144(k)^n) = n. - Ruediger Jehn, Jan 14 2022

A009000 Ordered hypotenuse numbers (squares are sums of 2 distinct nonzero squares).

Original entry on oeis.org

5, 10, 13, 15, 17, 20, 25, 25, 26, 29, 30, 34, 35, 37, 39, 40, 41, 45, 50, 50, 51, 52, 53, 55, 58, 60, 61, 65, 65, 65, 65, 68, 70, 73, 74, 75, 75, 78, 80, 82, 85, 85, 85, 85, 87, 89, 90, 91, 95, 97, 100, 100, 101, 102, 104, 105, 106, 109, 110, 111, 113, 115, 116, 117, 119, 120

Offset: 1

David W. Wilson

The largest member 'c' of the Pythagorean triples (a,b,c) ordered by increasing c.
If c^2 = a^2 + b^2 (a < b < c) then c^2 = (n^2 + m^2)/2 with n = b - a, m = b + a. - Zak Seidov, Mar 03 2011
Numbers n such that A083025(n) > 0, i.e., n is divisible by at least one prime of the form 4k+1. - Max Alekseyev, Oct 24 2008
A number appears only once in the sequence if and only if it is divisible by exactly one prime of the form 4k+1 with multiplicity one (cf. A084645). - Jean-Christophe Hervé, Nov 11 2013
If c^2 = a^2 + b^2 with a and b > 0, then a <> b: the sum of 2 equal squares cannot be a square because sqrt(2) is not rational. - Jean-Christophe Hervé, Nov 11 2013

W. L. Schaaf, Recreational Mathematics, A Guide To The Literature, "The Pythagorean Relationship", Chapter 6 pp. 89-99 NCTM VA 1963.
W. L. Schaaf, A Bibliography of Recreational Mathematics, Vol. 2, "The Pythagorean Relation", Chapter 6 pp. 108-113 NCTM VA 1972.
W. L. Schaaf, A Bibliography of Recreational Mathematics, Vol. 3, "Pythagorean Recreations", Chapter 6 pp. 62-6 NCTM VA 1973.

Zak Seidov and T. D. Noe, Table of n, a(n) for n = 1..10000 (Zak Seidov entered the first 1981 terms).
Anonymous, Links to Pythagorean Theorem Proofs
Henry Bottomley, Pythagoras's theorem (animated proof)
Kangourou Math Website, L'animation du theoreme de Pythagore
Ron Knott, Pythagorean Triples and Online Calculators
Ron Knott, Right-angled Triangles and Pythagoras' Theorem
I. Kobayashi et al., Pythagorean Theorem (Java Interactive Proofs, Applications and Explorations)
Mathematical Database, Poster, 7 Ways to prove the Pythagorean Theorem
B. Richmond, The Pythagorean Theorem
M. Shepperd, Web Resources on Pythagoras' Theorem
J. S. Silverman, A Friendly Introduction to Number Theory, Chapters 1 to 6 (see Chapters 2 and 3).
G. Villemin's Almanach of Numbers, Triangles & Triplets de Pythagore
Eric Weisstein's World of Mathematics, Pythagorean Triple
Index entries for sequences related to sums of squares

Cf. A009012, A009003, A024507, A004431, A046083, A046084, A004144, A083025, A084645, A084646, A084647, A084648, A084649, A006339.

A009000:=proc(N) # To get all terms <= N
    local p,q,i,L;
    L:=[];
    for p from 2 to floor(sqrt(N-1)) do
        for q to p-1 do
            if igcd(p,q)=1 and is(p-q,odd) then
                L:=[op(L),seq(i*(p^2+q^2),i=1..N/(p^2+q^2))];
            fi
        od
    od;
    return op(sort(L))
end proc:
A009000(120); # Felix Huber, Feb 10 2025

max = 120; hypotenuseQ[n_] := For[k = 1, True, k++, p = Prime[k]; Which[Mod[p, 4] == 1 && Divisible[n, p], Return[True], p > n, Return[False]]]; hypotenuses = Select[Range[max], hypotenuseQ]; red[c_] := {a, b, c} /. {ToRules[ Reduce[0 < a <= b && a^2 + b^2 == c^2, {a, b}, Integers]]}; A009000 = Flatten[red /@ hypotenuses, 1][[All, -1]] (* Jean-François Alcover, May 23 2012, after Max Alekseyev *)
Sqrt[#]&/@Flatten[Table[Total/@Select[IntegerPartitions[n^2,{2}],Length[Union[#]]==2&&AllTrue[Sqrt[#],IntegerQ]&],{n,150}]] (* Harvey P. Dale, May 25 2025 *)

list(lim)=my(v=List(),m2,s2,h2,h); for(middle=4,lim-1, m2=middle^2; for(small=1,middle, s2=small^2; if(issquare(h2=m2+s2,&h), if(h>lim, break); listput(v, h)))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 23 2017

list(lim) = {my(lh = List()); for(u = 2, sqrtint(lim), for(v = 1, u, if (u^2+v^2 > lim, break); if ((gcd(u,v) == 1) && (0 != (u-v)%2), for (i = 1, lim, if (i*(u^2+v^2) > lim, break); /* if (u^2 - v^2 < 2*u*v, w = [i*(u^2 - v^2), i*2*u*v, i*(u^2+v^2)], w = [i*2*u*v, i*(u^2 - v^2), i*(u^2+v^2)]); */ listput(lh, i*(u^2+v^2)););););); vecsort(Vec(lh));} \\ Michel Marcus, Apr 10 2021

from math import isqrt
def aupto(limit):
  s = [i*i for i in range(1, limit+1)]
  s2 = sorted(a+b for i, a in enumerate(s) for b in s[i+1:])
  return [isqrt(k) for k in s2 if k in s]
print(aupto(120)) # Michael S. Branicky, May 10 2021

A084647 Hypotenuses for which there exist exactly 3 distinct integer triangles.

Original entry on oeis.org

125, 250, 375, 500, 750, 875, 1000, 1125, 1375, 1500, 1750, 2000, 2197, 2250, 2375, 2625, 2750, 2875, 3000, 3375, 3500, 3875, 4000, 4125, 4394, 4500, 4750, 4913, 5250, 5375, 5500, 5750, 5875, 6000, 6125, 6591, 6750, 7000, 7125, 7375, 7750

Offset: 1

Eric W. Weisstein, Jun 01 2003

nonn

Numbers whose square is decomposable in 3 different ways into the sum of two nonzero squares: these are those with exactly one prime divisor of the form 4k+1 with multiplicity three. - Jean-Christophe Hervé, Nov 11 2013

			a(1) = 125 = 5^3, and 125^2 = 100^2 + 75^2 = 117^2 + 44^2 = 120^2 + 35^2. - _Jean-Christophe Hervé_, Nov 11 2013

Ray Chandler, Table of n, a(n) for n = 1..10000 (first 1140 terms from Jean-Christophe Hervé)
Eric Weisstein's World of Mathematics, Pythagorean Triple

Cf. A002144, A006339, A046080, A046109, A083025.

Cf. A004144 (0), A084645 (1), A084646 (2), A084648 (4), A084649 (5), A097219 (6), A097101 (7), A290499 (8), A290500 (9), A097225 (10), A290501 (11), A097226 (12), A097102 (13), A290502 (14), A290503 (15), A097238 (16), A097239 (17), A290504 (18), A290505 (19), A097103 (22), A097244 (31), A097245 (37), A097282 (40), A097626 (67).

Clear[lst,f,n,i,k] f[n_]:=Module[{i=0,k=0},Do[If[Sqrt[n^2-i^2]==IntegerPart[Sqrt[n^2-i^2]],k++ ],{i,n-1,1,-1}]; k/2]; lst={}; Do[If[f[n]==3,AppendTo[lst,n]],{n,4*5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Aug 12 2009 *)

Terms are obtained by the products A004144(k)*A002144(p)^3 for k, p > 0, ordered by increasing values. - Jean-Christophe Hervé, Nov 12 2013

A084648 Hypotenuses for which there exist exactly 4 distinct integer triangles.

Original entry on oeis.org

65, 85, 130, 145, 170, 185, 195, 205, 221, 255, 260, 265, 290, 305, 340, 365, 370, 377, 390, 410, 435, 442, 445, 455, 481, 485, 493, 505, 510, 520, 530, 533, 545, 555, 565, 580, 585, 595, 610, 615, 625, 629, 663, 680, 685, 689, 697, 715, 730, 740, 745

Offset: 1

Eric W. Weisstein, Jun 01 2003

nonn

Numbers whose square is decomposable in 4 different ways into the sum of two nonzero squares: these are those with exactly 2 distinct prime divisors of the form 4k+1, each with multiplicity one, or with only one prime divisor of this form with multiplicity 4. - Jean-Christophe Hervé, Nov 11 2013

If m is a term, then 2*m and p*m are terms where p is any prime of the form 4k+3. - Ray Chandler, Dec 30 2019

			a(1) = 65 = 5*13, and 65^2 = 52^2 + 39^2 = 56^2 + 33^2 = 60^2 + 25^2 = 63^2 + 16^2. - _Jean-Christophe Hervé_, Nov 11 2013

Ray Chandler, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Pythagorean Triple

Cf. A002144, A006339, A046080, A046109, A083025.

Cf. A004144 (0), A084645 (1), A084646 (2), A084647 (3), A084649 (5), A097219 (6), A097101 (7), A290499 (8), A290500 (9), A097225 (10), A290501 (11), A097226 (12), A097102 (13), A290502 (14), A290503 (15), A097238 (16), A097239 (17), A290504 (18), A290505 (19), A097103 (22), A097244 (31), A097245 (37), A097282 (40), A097626 (67).

Clear[lst,f,n,i,k] f[n_]:=Module[{i=0,k=0},Do[If[Sqrt[n^2-i^2]==IntegerPart[Sqrt[n^2-i^2]],k++ ],{i,n-1,1,-1}]; k/2]; lst={}; Do[If[f[n]==4,AppendTo[lst,n]],{n,6!}]; lst (* Vladimir Joseph Stephan Orlovsky, Aug 12 2009 *)

A084649 Hypotenuses for which there exist exactly 5 distinct Pythagorean triangles.

Original entry on oeis.org

3125, 6250, 9375, 12500, 18750, 21875, 25000, 28125, 34375, 37500, 43750, 50000, 56250, 59375, 65625, 68750, 71875, 75000, 84375, 87500, 96875, 100000, 103125, 112500, 118750, 131250, 134375, 137500, 143750, 146875, 150000, 153125

Offset: 1

Eric W. Weisstein, Jun 01 2003

nonn

Numbers whose square is decomposable in 5 different ways into the sum of two nonzero squares: these are those with exactly one prime divisor of the form 4k+1 with multiplicity 5. - Jean-Christophe Hervé, Nov 12 2013

			a(1) = 5^5, a(5) = 6*5^5, a(65) = 13^5. - _Jean-Christophe Hervé_, Nov 12 2013

Ray Chandler, Table of n, a(n) for n = 1..10000 (first 1019 terms from Jean-Christophe Hervé)
Eric Weisstein's World of Mathematics, Pythagorean Triple

Cf. A002144, A006339, A046080, A046109, A083025.

Cf. A004144 (0), A084645 (1), A084646 (2), A084647 (3), A084648 (4), A097219 (6), A097101 (7), A290499 (8), A290500 (9), A097225 (10), A290501 (11), A097226 (12), A097102 (13), A290502 (14), A290503 (15), A097238 (16), A097239 (17), A290504 (18), A290505 (19), A097103 (22), A097244 (31), A097245 (37), A097282 (40), A097626 (67).

Clear[lst,f,n,i,k] f[n_]:=Module[{i=0,k=0},Do[If[Sqrt[n^2-i^2]==IntegerPart[Sqrt[n^2-i^2]],k++ ],{i,n-1,1,-1}]; k/2]; lst={}; Do[If[f[n]==5,AppendTo[lst,n]],{n,3*6!}]; lst (* Vladimir Joseph Stephan Orlovsky, Aug 12 2009 *)

Terms are obtained by the products A004144(k)*A002144(p)^5 for k, p > 0 ordered by increasing values. - Jean-Christophe Hervé, Nov 12 2013

cp's OEIS Frontend

A350056 Perimeter of integer-sided triangle with hypotenuse A084645(n).

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A350057 Short leg lengths of integer-sided triangles with hypotenuse A084645(n).

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A350058 Long leg lengths of integer-sided triangles with hypotenuse A084645(n).

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A350564 Areas of integer-sided triangles with hypotenuse A084645(n).

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A009003 Hypotenuse numbers (squares are sums of 2 nonzero squares).

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A046080 a(n) is the number of integer-sided right triangles with hypotenuse n.

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A009000 Ordered hypotenuse numbers (squares are sums of 2 distinct nonzero squares).

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Maple

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A084647 Hypotenuses for which there exist exactly 3 distinct integer triangles.

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