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Let a(1)=3; for n > 1, let a(n) be the least positive integer k such that k > a(n-1), a(1)^2 + ... + a(n-1)^2 + k^2 is a square and the Pythagorean triple sqrt(a(1)^2 + ... + a(n-1)^2), a(n), sqrt(a(1)^2 + ... + a(n)^2) is primitive.
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%I #67 Feb 16 2023 21:33:15

%S 3,4,12,84,132,12324,89892,2447844,28350372,295742791596,

%T 171480834409712412,633511848768467916,

%U 1616599508725767821225590810932,4158520496012961741299012805876,115366949386695884000892071516523067413910188

%N Let a(1)=3; for n > 1, let a(n) be the least positive integer k such that k > a(n-1), a(1)^2 + ... + a(n-1)^2 + k^2 is a square and the Pythagorean triple sqrt(a(1)^2 + ... + a(n-1)^2), a(n), sqrt(a(1)^2 + ... + a(n)^2) is primitive.

%C For n > 1, a(n) is the even value of a primitive Pythagorean triple where the larger odd value of the triple equals the smaller odd value of a primitive Pythagorean triple with even value a(n+1) (see A239381). - _Torlach Rush_, Jan 27 2023

%F The numbers are generated by using the well-known characterization of primitive Pythagorean triples, namely (a,b,c) is a PPT iff there are positive integers j,k of opposite parity with j > k, and gcd(j,k)=1 such that a = j^2 - k^2, b = 2jk, c = j^2 + k^2.

%o (PARI) lista(NN) = s=9;k=3;print1(k);for(n=1,NN-1,v=divisors(s);j=#v;while(v[j]*(v[j]+2*k)>s,j--);while(gcd((s-v[j]^2)/(2*v[j]), s)!=1, j--);print1(", ", k=(s-v[j]^2)/(2*v[j]));s+=k^2); \\ _Jinyuan Wang_, May 31 2019

%Y Cf. A018930, A127689, A239381.

%K nonn

%O 1,1

%A _Rohan Hemasingha_, May 30 2019

%E a(14)-a(15) from _Jinyuan Wang_, Jun 01 2019