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A297294
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Number of primitive Pythagorean triples (PPTs) that have 2^n-1 as the length of their odd leg where n is the sequence index.
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1
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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
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1,4
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COMMENTS
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2^n-1 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^n-1 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^n-1, x < y and gcd(x, y) = 1. Using Euclid's parametric generators for PPTs (s^2+t^2, 2s*t, s^2-t^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^n-1 as the length of the odd leg providing that s = (x+y)/2 and t = (y-x)/2.
If 2^n-1 has d distinct prime factors (A046800(n)), then the set of pairs {x, y} such that x*y = 2^n-1, x < y and gcd(x, y) = 1 has a cardinality of 2^(d-1). 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^(d-1). Let m be the product of the distinct factor of 2^n-1 and r be the remainder consisting of the remaining repeated prime factors where m*r = 2^n-1. 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^n-1 and {x, y} created from all the prime factors of 2^n-1 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.
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LINKS
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FORMULA
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For n=1, a(n)=0 otherwise a(n)=2^(A046800(n)-1).
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EXAMPLE
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a(6)=2, because 2^6-1 = 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.
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MATHEMATICA
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pairs[n_] := Module[{m=2^n-1, lst=Divisors[2^n-1]}, 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^n-1]-1)]; Array[a, 100]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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