%I #40 Mar 22 2021 15:03:25

%S 1,1,4,4,9,12,23,2,19,46,67,74,86,109,122,64,103,167,191,236,281,292,

%T 359,449,512,568,601,607,664,673,743,59,132,531,581,627,876,1008,1284,

%U 1588,1659,1723,2092,2136,2317,2373,2736,2757,2803,3072,3164,3333,3469,3704,3821,4028,4077,4136,4371,4596,4668,4712,4851

%N Irregular triangle read by rows: row n consists of all numbers x such that x^2 + y^2 = A006278(n), with 0 < x < y.

%C 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.

%H Andrew Howroyd, <a href="/A341678/b341678.txt">Table of n, a(n) for n = 1..1023</a> (rows 1..10)

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SumofSquaresFunction.html">Sum of Squares Function</a>

%e Triangle starts:

%e 1,

%e 1, 4,

%e 4, 9, 12, 23,

%e 2, 19, 46, 67, 74, 86, 109, 122,

%e 64, 103, 167, 191, 236, 281, 292, 359, 449, 512, 568, 601, 607, 664, 673, 743,

%e ...

%e 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.

%o (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

%Y Cf. A006278, A230710.

%Y Cf. A236381 (1st column).

%K nonn,tabf

%O 1,3

%A _Richard Peterson_, Feb 17 2021