Proof for OEIS A373225
Michael S. Branicky, May 30 2024

Remark:
When p == 3 mod 4 is a term in A096448, it is a term of A373225 that is > 2. And vice versa.

Proof of ⇒:

Let p_n denote the nth prime. From A373223, we have

T(n, k) = { 0, if n = k;
    			{ 1, if n = 1 or k = 1;
    			{ (-1)^(((p_n - 1)/2)*((p_k - 1)/2)), o.w.

When p = p_n > 2, the row sum is

      Sum_{k=1..n} T(n, k) = T(n, 1) + T(n, 2) + ... + T(n, n-1) + T(n, n)
                           = 1   + T(n, 2) + ... + T(n, n-1) + 0
                           = 1 + [(-1)^(((p_n-1)/2)*((p_2-1)/2))) + ... + (-1)^(((p_n - 1)/2)*((p_{n-1} - 1)/2)))] (*)

When p_n = 3 mod 4, ((p_k-1)/2) is odd and the sign of each term is determined by whether (p_k-1)/2 is even or odd. 
When p_k = 1 mod 4, (p_k-1)/2 is even, so that term of the sum is 1;
when p_k = 3 mod 4, (p_k-1)/2 is odd, so that term of the sum is -1.

Now, suppose further that p_n is a term of A096448 ("Primes p such that the number of primes less than p equal to 1 mod 4 is one less than the number of primes less than p equal to 3 mod 4.")

Then, the sum has one less 1 than -1, so the sum in brackets is -1 and (*) is 0. #

Proof of ⇐:

When p_n == 1 mod 4, ((p_n-1)/2) is even, so every term in the brackets is 1 and (*) cannot be 0.  Thus, p_n == 3 mod 4 for terms > 2 in A373225.  So, assume p_n == 3 mod 4, thus ((p_n-1)/2) is odd.  Now, for (*) to be 0, we need the term in brackets to be -1. Since each term is -1 or 1, we need the number of -1's to be 1 more than the number of 1's, which happens when p_n is a member of A096448. #