A046081 Number of integer-sided right triangles with n as a hypotenuse or leg.
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
Keywords
Examples
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)
References
- A. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 116-117, 1966.
Links
- Lars Blomberg, Table of n, a(n) for n = 1..10000
- Anonymous, Generator of all Pythagorean triples that include a given number [Internet Archive Wayback Machine]
- 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 8.
- Eric Weisstein's World of Mathematics, Pythagorean Triple
Programs
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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 *)
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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
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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
Formula
Extensions
Improved name by Bernard Schott, Jan 03 2019
Comments