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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.
0
3, 4, 12, 84, 132, 12324, 89892, 2447844, 28350372, 295742791596, 171480834409712412, 633511848768467916, 1616599508725767821225590810932, 4158520496012961741299012805876, 115366949386695884000892071516523067413910188
OFFSET
1,1
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
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.
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Rohan Hemasingha, May 30 2019
EXTENSIONS
a(14)-a(15) from Jinyuan Wang, Jun 01 2019
STATUS
approved