A330487 Sum of those x(p)*y(p) with p <= n, where p is a prime congruent to 1 modulo 4, and p = x(p)^2 + y(p)^2 with 1 <= x(p) <= y(p).

0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2, 8, 8, 8, 8, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 22, 22, 22, 22, 22, 22, 22, 22, 28, 28, 28, 28, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 62, 62, 62, 62, 62, 62, 62, 62, 92, 92, 92, 92, 92, 92, 92, 92, 92, 92, 92, 92, 116, 116, 116, 116, 116, 116, 116, 116

Offset

1,5

Comments

Conjecture: Let b(n) be the sum of all primes p <= n with p == 1 (mod 4). Then a(n)/b(n) = 1/Pi + O(1/sqrt(n)).

A classical theorem of Euler (conjectured by Fermat) states that any prime p == 1 (mod 4) can be written uniquely as x^2 + y^2 with 1 <= x <= y.

For any prime p == 1 (mod 4), we obviously have a(p) > a(p-1). Also, b(n) >= 2*a(n) for all n > 0 since x^2 + y^2 >= 2*x*y.

Via computation we find that a(10^10) = 353452066546904620, b(10^10) = 1110397615780409147, and 3.14157907 < b(10^10)/a(10^10) < 3.14157908.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, A mysterious connection between prime and Pi, Question 348448 at MathOverFlow, December 15, 2019.

Zhi-Wei Sun, New series for powers of Pi and related congruences, arXiv:1911.05456 [math.NT], 2019.

Example

a(5) = 2 since 5 is the first prime congruent to 1 mod 4 and 5 = 1^2 + 2^2 with 1*2 = 2.
a(13) = 8 since 13 = 2^2 + 3^2 is the second prime congruent to 1 mod 4 and 1*2 + 2*3 = 8.

Mathematica

tab={}; Do[m=0; Do[If[PrimeQ[(2x+1)^2+(2y)^2], m=m+(2x+1)*(2y)], {x, 0, (Sqrt[n]-1)/2}, {y, 1, Sqrt[n-(2x+1)^2]/2}]; tab=Append[tab, m], {n, 1, 80}]; Print[tab]

Crossrefs

Cf. A000796, A002144, A002972, A002973.

Sequence in context: A365882 A103795 A123627 * A339169 A171818 A248777

Adjacent sequences: A330484 A330485 A330486 * A330488 A330489 A330490

Keyword

nonn

Author

Zhi-Wei Sun, Dec 15 2019

Status

approved