OFFSET
1,3
COMMENTS
The n-th row of the triangle is of length 2^(n-1), since a product of n distinct primes congruent to 1 (mod 4) has 2^(n-1) solutions to being the sum of two squares.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1023 (rows 1..10)
Eric Weisstein's World of Mathematics, Sum of Squares Function
EXAMPLE
Triangle starts:
1,
1, 4,
4, 9, 12, 23,
2, 19, 46, 67, 74, 86, 109, 122,
64, 103, 167, 191, 236, 281, 292, 359, 449, 512, 568, 601, 607, 664, 673, 743,
...
In the second row, calculations are as follows. 5*13 is the product of the first two primes congruent to 1 (mod 4), and 65 = 1^2 + 8^2 = 4^2 + 7^2, so the second row is 1, 4.
PROG
(PARI) row(n) = {my(t=1, q=3, v=vector(2^n/2)); for(k=1, n, until(q%4==1, q=nextprime(q+1)); t*=q); q=0; for(k=1, #v, until(issquare(t-q^2), q++); v[k]=q); v; } \\ Jinyuan Wang, Mar 03 2021
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Richard Peterson, Feb 17 2021
STATUS
approved