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A046081
Number of integer-sided right triangles with n as a hypotenuse or leg.
23
0, 0, 1, 1, 2, 1, 1, 2, 2, 2, 1, 4, 2, 1, 5, 3, 2, 2, 1, 5, 4, 1, 1, 7, 4, 2, 3, 4, 2, 5, 1, 4, 4, 2, 5, 7, 2, 1, 5, 8, 2, 4, 1, 4, 8, 1, 1, 10, 2, 4, 5, 5, 2, 3, 5, 7, 4, 2, 1, 14, 2, 1, 7, 5, 8, 4, 1, 5, 4, 5, 1, 12, 2, 2, 9, 4, 4, 5, 1, 11, 4, 2, 1, 13, 8, 1, 5, 7, 2, 8, 5, 4, 4, 1, 5, 13, 2, 2, 7
OFFSET
1,5
COMMENTS
Pythagorean triples including primitive ones and non-primitive ones. For a certain n, it may be a leg or the hypotenuse in either a primitive Pythagorean triple, or a non-primitive Pythagorean triple, or both. - Rui Lin, Nov 02 2019
REFERENCES
A. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 116-117, 1966.
LINKS
Anonymous, Generator of all Pythagorean triples that include a given number [Internet Archive Wayback Machine]
A. Tripathi, On Pythagorean triples containing a fixed integer, Fib. Q., 46/47 (2008/2009), 331-340. See Theorem 8.
Eric Weisstein's World of Mathematics, Pythagorean Triple
FORMULA
a(n) = A046079(n) + A046080(n). - Lekraj Beedassy, Dec 01 2003
From Rui Lin, Nov 02 2019: (Start)
a(n) = A024363(n) + A328949(n).
a(n) = A024361(n) + A024362(n) + A328708(n) + A328712(n). (End)
EXAMPLE
From Rui Lin, Nov 02 2019: (Start)
n=25 is the least number which meets all of following cases:
1. 25 is a leg of a primitive Pythagorean triple (25,312,313), so A024361(25)=1;
2. 25 is the hypotenuse of a primitive Pythagorean triple (7,24,25), so A024362(25)=1;
3. 25 is a leg of a non-primitive Pythagorean triple (25,60,65), so A328708(25)=1;
4. 25 is the hypotenuse of a non-primitive Pythagorean triple (15,20,25), so A328712(25)=1;
5. Combination 1. and 3. means A046079(25)=2;
6. Combination 2. and 4. means A046080(25)=2;
7. Combination 1. and 2. means A024363(25)=2;
8. Combination 3. and 4. means A328949(25)=2;
9. Combination of 1., 2., 3., and 4. means A046081(25)=4. (End)
MATHEMATICA
a[1] = 0; a[n_] := Module[{f}, f = Select[FactorInteger[n], Mod[#[[1]], 4] == 1&][[All, 2]]; (DivisorSigma[0, If[OddQ[n], n, n/2]^2]-1)/2 + (Times @@ (2*f+1) - 1)/2]; Array[a, 99] (* Jean-François Alcover, Jul 19 2017 *)
PROG
(PARI) a(n) = {oddn = n/(2^valuation(n, 2)); f = factor(oddn); for (k=1, #f~, if ((f[k, 1] % 4) != 1, f[k, 2] = 0); ); n1 = factorback(f); if (n % 2, (numdiv(n^2)+numdiv(n1^2))/2 -1, (numdiv((n/2)^2)+numdiv(n1^2))/2 -1); } \\ Michel Marcus, Mar 07 2016
(Python)
from sympy import factorint
def a(n):
p1, p2 = 1, 1
for i in factorint(n).items():
if i[0] % 4 == 1:
p2 *= i[1] * 2 + 1
p1 *= i[1] * 2 + 1 - (2 if i[0] == 2 else 0)
return (p1 + p2)//2 - 1
print([a(n) for n in range(1, 100)]) # Oleg Sorokin, Mar 02 2023
KEYWORD
nonn
EXTENSIONS
Improved name by Bernard Schott, Jan 03 2019
STATUS
approved