A132404 -id:A132404 - cp's OEIS frontend

A024359 Number of primitive Pythagorean triangles with short leg n.

0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 0, 1, 1, 1, 0, 1, 2, 1, 0, 1, 1, 2, 0, 1, 2, 1, 0, 2, 1, 1, 0, 1, 2, 1, 0, 1, 2, 1, 0, 2, 2, 1, 0, 1, 1, 2, 0, 1, 3, 1, 0, 1, 1, 2, 0, 1, 2, 2, 0, 1, 1, 1, 0, 2, 2, 1, 0, 1, 1, 1, 0, 1, 3, 2, 0, 2

Offset: 1

David W. Wilson

nonn

Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence gives number of times A takes value n.
Number of times n occurs in A020884.
a(A139544(n)) = 0; a(A024352(n)) > 0. - Reinhard Zumkeller, Nov 09 2012

T. D. Noe, Table of n, a(n) for n = 1..10000
Ron Knott, Pythagorean Triples and Online Calculators

Cf. A020884, A024352, A024360, A024361, A132404 (where records occur), A139544.

a024359_list = f 0 1 a020884_list where
   f c u vs'@(v:vs) | u == v = f (c + 1) u vs
                    | u /= v = c : f 0 (u + 1) vs'
-- Reinhard Zumkeller, Nov 09 2012

solns[a_] := Module[{b, c, soln}, soln = Reduce[a^2 + b^2 == c^2 && a < b && c > 0 && GCD[a, b, c] == 1, {b, c}, Integers]; If[soln === False, 0, If[soln[[1, 1]] === b, 1, Length[soln]]]]; Table[solns[n], {n, 100}]
(* Second program: *)
a[n_] := Module[{s = 0, b, c, d, g}, Do[g = Quotient[n^2, d]; If[d <= g && Mod[d+g, 2] == 0, c = Quotient[d+g, 2]; b = g-c; If[n < b && GCD[b, c] == 1, s++]], {d, Divisors[n^2]}]; s]; Array[a, 100] (* Jean-François Alcover, Apr 27 2019, from PARI *)

nppt(a) = {
  my(s=0, b, c, d, g);
  fordiv(a^2, d,
    g=a^2\d;
    if(d<=g && (d+g)%2==0,
      c=(d+g)\2; b=g-c;
      if(aColin Barker, Oct 25 2015

Formula

a(n) = A024361(n) - A024360(n). - Ray Chandler, Feb 03 2020

A307706 Number of unitary divisors of n that are smaller than sqrt((sqrt(2) - 1)*n).

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 3, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1, 2, 2, 1, 3, 1, 2, 1, 1, 2, 3, 1, 2, 2, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 1, 2, 1, 3, 2, 2, 2, 1, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 1, 2, 1, 4, 1

Offset: 1

Jianing Song, Apr 23 2019

nonn

Related to A024359: (Start)
Note that all the primitive Pythagorean triangles are given by A = min{2*u*v, u^2 - v^2}, B = max{2*u*v, u^2 - v^2}, C = u^2 + v^2, where u, v are coprime positive integers, u > v and u - v is odd. As a result:
(a) if n is odd, then A024359(n) is the number of representations of n to the form n = u^2 - v^2, where u, v are coprime positive integers (note that this guarantees that u - v is odd), u > v and u^2 - v^2 < 2*u*v. Let s = u + v, t = u - v, then n = s*t, where s and t are unitary divisors of n, s > t and s*t < (s^2 - t^2)/2, so t is in the range (0, sqrt((sqrt(2) - 1)*n));
(b) if n is divisible by 4, then A024359(n) is the number of representations of n to the form n = 2*u*v, where u, v are coprime positive integers (note that this also guarantees that u - v is odd because n/2 is even), u > v and 2*u*v < u^2 - v^2. So u and v must be unitary divisors of n/2, and v is in the range (0, sqrt((sqrt(2) - 1)*(n/2))).
(c) if n == 2 (mod 4), then n/2 is odd, so n = 2*u*v implies that u and v are both odd, which is not acceptable, so A024359(n) = 0.
Similarly, let b(n) be the number of unitary divisors of n in the range (sqrt((sqrt(2) - 1)*n), sqrt(n)) (= A034444(n)/2 - a(n) for n > 1), then the number of times B takes value n is b(n) for odd n > 1, b(n/2) if n is divisible by 4 and 0 if n = 1 or n == 2 (mod 4). (End)
For k >= 2, the earliest occurrence of k is at n = A132404(k)/2 if A132404(k) is even (and thus being a multiple of 4). Conjecture: this is always the case.

			The unitary divisors of 210 that are smaller than sqrt((sqrt(2) - 1)*210) = 9.3265... are 1, 2, 3, 5, 6 and 7, so a(210) = 6. Correspondingly, A024359(420) = 6.

Ray Chandler, Table of n, a(n) for n = 1..10000

Cf. A024359, A034444, A132404.

a(n) = my(i=0); for(k=1, sqrt((sqrt(2)-1)*n), if(!(n%k) && gcd(k,n/k)==1, i++)); i

A024359(n) = a(n) for odd n; A024359(n) = a(n/2) for n divisible by 4.

A379596 a(n) is the least positive integer k for which k^2 + (k + n)^2 is a square.

Original entry on oeis.org

3, 6, 9, 12, 15, 18, 5, 24, 27, 30, 33, 36, 39, 10, 45, 48, 7, 54, 57, 60, 15, 66, 12, 72, 75, 78, 81, 20, 87, 90, 9, 96, 99, 14, 25, 108, 111, 114, 117, 120, 36, 30, 129, 132, 135, 24, 16, 144, 11, 150, 21, 156, 159, 162, 165, 40, 171, 174, 177, 180, 183, 18, 45

Offset: 1

Felix Huber, Feb 15 2025

a(n) is also the smallest short leg of a Pythagorean triangle where the difference between the two legs is n.

A289398(n) is the least integer m > n for which (n^2 + m^2)/2 is a square. This is equivalent to the least positive integer k for which (n^2 + (n + 2*k)^2)/2 = k^2 + (n + k)^2 is a square. From m = n + 2*k follows a(n) = (A289398(n) - n)/2.

			a(1) = 3 because 3^2 + (3 + 1)^2 = 5^2 and there is no smaller positive integer k than 3 with that property.
a(28) = 20 because 20^2 + (20 + 28)^2 = 52^2 and there is no smaller positive integer k than 20 with that property.

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

Cf. A089015, A009004, A120682, A132404, A289398, A379830.

A379596:=proc(n)
    local k;
    for k do
        if issqr(k^2+(k+n)^2) then
            return k
        fi
    od
end proc;
seq(A379596(n),n=1..63);

s={};Do[k=0;Until[IntegerQ[Sqrt[k^2+(k+n)^2]],k++];AppendTo[s,k],{n,63}];s (* James C. McMahon, Mar 02 2025 *)

a(n) = my(k=1); while (!issquare(k^2 + (k + n)^2), k++); k; \\ Michel Marcus, Feb 15 2025

from itertools import count
from sympy.ntheory.primetest import is_square
def A379596(n): return next(k for k in count(1) if is_square(k**2+(k+n)**2)) # Chai Wah Wu, Mar 02 2025

a(n) = (A289398(n) - n)/2.

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A024359 Number of primitive Pythagorean triangles with short leg n.

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A307706 Number of unitary divisors of n that are smaller than sqrt((sqrt(2) - 1)*n).

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A379596 a(n) is the least positive integer k for which k^2 + (k + n)^2 is a square.

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