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

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

1,25

Comments

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

References

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

Links

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

Formula

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

Maple

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

Mathematica

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 *)

Prog

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

Crossrefs

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.

Sequence in context: A088950 A267113 A083025 * A170967 A035227 A355320

Adjacent sequences: A046077 A046078 A046079 * A046081 A046082 A046083

Keyword

nonn

Author

Eric W. Weisstein

Status

approved