|
|
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).
|
|
1
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
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
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|