OFFSET
1,8
COMMENTS
Number of ways in which n can be the leg (other than the hypotenuse) of a primitive or nonprimitive right triangle.
Number of ways that 2/n can be written as a sum of exactly two distinct unit fractions. For every solution to 2/n = 1/x + 1/y, x < y, the Pythagorean triple is (n, y-x, x+y-n). - T. D. Noe, Sep 11 2002
For n>2, the positions of the ones in this sequence correspond to the prime numbers and their doubles, A001751. - Ant King, Jan 29 2011
Let L = length of longest leg, H = hypotenuse. For odd n: L =(n^2-1)/2 and H = L+1. For even n, L = (n^2-4)/4 and H = L+2. - Richard R. Forberg, May 31 2013
Or number of ways n^2 can be written as the difference of two positive squares: a(3) = 1: 3^2 = 5^2-4^2; a(8) = 2: 8^2 = 10^2-6^2 = 17^2-15^2; a(16) = 3: 16^2 = 20^2-12^2 = 34^2-30^2 = 65^2-63^2. - Alois P. Heinz, Aug 06 2019
Number of ways to write 2n as the sum of two positive integers r and s such that r < s and (s - r) | (s * r). - Wesley Ivan Hurt, Apr 21 2020
REFERENCES
Albert H. Beiler, Recreations in the Theory of Numbers. New York: Dover Publications, 1966, pp. 116-117.
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..65537
Hanz Becker, Pythagorean triples in JavaScript.
Ron Knott, Pythagorean Triples and Online Calculators.
Project Euler, Problem 176: Right-angled triangles that share a cathetus.
Fred Richman, Pythagorean Triples. [Wayback Machine link]
Amitabha Tripathi, On Pythagorean triples containing a fixed integer, Fib. Q., 46/47 (2008/2009), 331-340. See Theorem 6.
Eric Weisstein's World of Mathematics, Pythagorean Triple.
FORMULA
For odd n, a(n) = A018892(n) - 1.
Let n = (2^a0)*(p1^a1)*...*(pk^ak). Then a(n) = [(2*a0 - 1)*(2*a1 + 1)*(2*a2 + 1)*(2*a3 + 1)*...*(2*ak + 1) - 1]/2. Note that if there is no a0 term, i.e., if n is odd, then the first term is simply omitted. - Temple Keller (temple.keller(AT)gmail.com), Jan 05 2008
For odd n, a(n) = (tau(n^2) - 1) / 2; for even n, a(n) = (tau((n / 2)^2) - 1) / 2. - Amber Hu (hupo001(AT)gmail.com), Jan 23 2008
a(n) = Sum_{i=1..n-1} (1 - ceiling(i*(2*n-i)/(2*n-2*i)) + floor(i*(2*n-i)/(2*n-2*i))). - Wesley Ivan Hurt, Apr 21 2020
Sum_{k=1..n} a(k) ~ (n / Pi^2) * (log(n)^2 + c_1 * log(n) + c_2), where c_1 = 2 * (gamma - 1) + 48*log(A) - 4*log(Pi) - 13*log(2)/3 = 3.512088... (gamma = A001620, log(A) = A225746), and c_2 = 6 * gamma^2 - (6 + log(2)) * gamma + 2 - Pi^2/2 + 19*log(2)^2/18 + log(2)/3 - 6*gamma_1 + 8 * (zeta'(2)/zeta(2))^2 + (4 - 12*gamma + 2*log(2)/3) * zeta'(2)/zeta(2) - 4*zeta''(2)/zeta(2) = -4.457877... (gamma_1 = -A082633). - Amiram Eldar, Nov 08 2024
MATHEMATICA
a[n_] := (DivisorSigma[0, If[OddQ[n], n, n / 2]^2] - 1) / 2; Table[a[i], {i, 100}] (* Amber Hu (hupo001(AT)gmail.com), Jan 23 2008 *)
a[ n_] := Length @ FindInstance[ n > 0 && y > 0 && z > 0 && n^2 + y^2 == z^2, {y, z}, Integers, 10^9]; (* Michael Somos, Jul 25 2018 *)
PROG
(Sage) def A046079(n) : return (number_of_divisors(n^2 if n%2==1 else n^2/4) - 1) // 2 # Eric M. Schmidt, Jan 26 2013
(PARI) A046079(n) = ((numdiv(if(n%2, n, n/2)^2)-1)/2); \\ Antti Karttunen, Sep 27 2018
(Python)
from math import prod
from sympy import factorint
def A046079(n): return prod((e+(p&1)<<1)-1 for p, e in factorint(n).items())>>1 # Chai Wah Wu, Sep 06 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved