A371497 - OEIS

%I #15 Apr 20 2024 10:16:55

%S 1,1,1,1,3,9,1,5,7,9,19,1,3,4,1,2,4,8,1,3,5,9,15,17,25,27,31,1,7,9,11,

%T 13,15,19,25,29,41,43,1,4,5,6,7,9,13,1,3,5,9,11,15,23,25,27,33,41,43,

%U 45,49,55,1,3,4,7,9,10,11,12,16,1,2,4,5,8,9,10,16,18,20,1,3,7,9,13,17

%N Irregular triangle read by rows: n-th row gives congruence classes s such that the n-th prime q is a quadratic residue modulo an odd prime p if and only if p = plus or minus s for some s (mod m), where m = q if q is of the form 4k + 1, else m = 4q.

%C If n-th prime q is of the form 4k + 1, then by quadratic reciprocity row n consists of quadratic residues mod q, that are less than 2k; i.e., for q > 3, the first half of the corresponding row in A063987.

%C The first term in each row is always 1.

%e The 1st prime, 2, not of the form 4k + 1, is a square modulo odd primes p if and only if p = +/- 1 (mod 4*2 = 8).

%e The 6th prime, 13, of the form 4k + 1, is a square modulo odd primes p if and only if p = +/- 1, +/- 3, or +/- 4 (mod 13).

%e The irregular triangle T(n,k) begins (q is prime(n)):

%e  n   q \k 1 2 3 4 5 6 7 8 9 10 11

%e  1,  2: 1

%e  2,  3: 1

%e  3,  5: 1

%e  4,  7: 1 3 9

%e  5, 11: 1 5 7 9 19

%e  6: 13: 1 3 4

%e  7, 17: 1 2 4 8

%e  8, 19: 1 3 5 9 15 17 25 27 31

%e  9, 23: 1 7 9 11 13 15 19 25 29 41 43

%e 10, 29: 1 4 5 6 7 9 13

%o (Python)

%o from sympy import prime

%o def A371497_row(n):

%o     q = prime(n)

%o     res = {i*i % q for i in range(1, q//2 + 1)}

%o     if q % 4 == 1:

%o         res = {a for a in res if 2*a < q}

%o     else:

%o         res = {((a % 4 - 1) * q + a) % (4*q) for a in res}

%o         res = {a if a < 2*q else 4*q - a for a in res}

%o     return sorted(res)

%Y Cf. A063987, A081728.

%K nonn,tabf,easy

%O 1,5

%A _Nick Hobson_, Mar 25 2024