A307077 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.

3, 4, 12, 84, 132, 12324, 89892, 2447844, 28350372, 295742791596, 171480834409712412, 633511848768467916, 1616599508725767821225590810932, 4158520496012961741299012805876, 115366949386695884000892071516523067413910188

Offset: 1

Author

Rohan Hemasingha, May 30 2019

Keywords

• nonn

Comments

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

Crossrefs

Cf. A018930, A127689, A239381.

Programs

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

Formula

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.

Extensions

a(14)-a(15) from Jinyuan Wang, Jun 01 2019