

A297294


Number of primitive Pythagorean triples (PPTs) that have 2^n1 as the length of their odd leg where n is the sequence index.


1



0, 1, 1, 2, 1, 2, 1, 4, 2, 4, 2, 8, 1, 4, 4, 8, 1, 8, 1, 16, 4, 8, 2, 32, 4, 4, 4, 32, 4, 32, 1, 16, 8, 4, 8, 128, 2, 4, 8, 64, 2, 32, 4, 64, 32, 8, 4, 256, 2, 64, 16, 64, 4, 32, 32, 128, 8, 32, 2, 1024, 1, 4, 32, 64, 4, 128, 2, 64, 8, 256, 4, 2048, 4, 16, 64, 64, 8, 64, 4, 256, 32, 16, 2, 2048, 4, 16, 32, 512, 1, 1024
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OFFSET

1,4


COMMENTS

2^n1 for n = 0 and 1 give the Mersenne numbers 0 and 1, neither of which can be the side length of a PPT. For n > 1, all Mersenne numbers are congruent to 3 mod 4. Consequently, no Mersenne number can be the length of the hypotenuse of a PPT.
If 2^n1 is the length of the odd leg of a PPT its divisors can provide a set of pairs {x, y} such that for each pair, x*y = 2^n1, x < y and gcd(x, y) = 1. Using Euclid's parametric generators for PPTs (s^2+t^2, 2s*t, s^2t^2) with s > t > 0 as positive integers, gcd(s, t) = 1 and s+t odd it is possible to generate all PPTs with 2^n1 as the length of the odd leg providing that s = (x+y)/2 and t = (yx)/2.
If 2^n1 has d distinct prime factors (A046800(n)), then the set of pairs {x, y} such that x*y = 2^n1, x < y and gcd(x, y) = 1 has a cardinality of 2^(d1). This is because an integer m consisting of d distinct factors will have 2^d divisors and will generate pairs {x', y'} such that x'*y' = m, x' < y' and gcd(x', y') = 1 with a cardinality of 2^(d1). Let m be the product of the distinct factor of 2^n1 and r be the remainder consisting of the remaining repeated prime factors where m*r = 2^n1. Then there has to be a 1 to 1 correspondence between the set of pairs {x', y'} created from the distinct prime factors of 2^n1 and {x, y} created from all the prime factors of 2^n1 whenever the repeated prime factors of r are combined with the distinct factors of m in the pairs {x, y} in order to preserve gcd(x, y) = 1.


LINKS

Frank M Jackson, Table of n, a(n) for n = 1..928


FORMULA

For n=1, a(n)=0 otherwise a(n)=2^(A046800(n)1).


EXAMPLE

a(6)=2, because 2^61 = 63 gives pairs {1, 63}, {3, 21}, {7, 9} whose members when multiplied give 63. However, only two of these pairs are coprime and will generate PPTs.


MATHEMATICA

pairs[n_] := Module[{m=2^n1, lst=Divisors[2^n1]}, Table[{lst[[l]], m/lst[[l]]}, {l, 1, Length[lst]/2}]]; Table[Length@Select[pairs[n], GCD@@#==1 &], {n, 1, 100}]
a[n_] := If[n==1, 0, 2^(Length@FactorInteger[2^n1]1)]; Array[a, 100]


CROSSREFS

Cf. A001265, A046800.
Sequence in context: A249029 A075997 A244517 * A161309 A161243 A161028
Adjacent sequences: A297291 A297292 A297293 * A297295 A297296 A297297


KEYWORD

nonn


AUTHOR

Frank M Jackson, Jan 04 2018


STATUS

approved



