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A046109
Number of lattice points (x,y) on the circumference of a circle of radius n with center at (0,0).
31
1, 4, 4, 4, 4, 12, 4, 4, 4, 4, 12, 4, 4, 12, 4, 12, 4, 12, 4, 4, 12, 4, 4, 4, 4, 20, 12, 4, 4, 12, 12, 4, 4, 4, 12, 12, 4, 12, 4, 12, 12, 12, 4, 4, 4, 12, 4, 4, 4, 4, 20, 12, 12, 12, 4, 12, 4, 4, 12, 4, 12, 12, 4, 4, 4, 36, 4, 4, 12, 4, 12, 4, 4, 12, 12, 20, 4, 4, 12, 4, 12, 4, 12, 4, 4, 36
OFFSET
0,2
COMMENTS
Also number of Gaussian integers x + yi having absolute value n. - Alonso del Arte, Feb 11 2012
LINKS
Eric Weisstein's World of Mathematics, Circle Lattice Points
FORMULA
a(n) = A000328(n) - A051132(n).
a(n) = 8*A046080(n) + 4 for n > 0.
a(n) = A004018(n^2).
From Jean-Christophe Hervé, Dec 01 2013: (Start)
a(A084647(k)) = 28.
a(A084648(k)) = 36.
a(A084649(k)) = 44. (End)
a(n) = 4 * Product_{i=1..k} (2*e_i + 1) for n > 0, given that p_i^e_i is the i-th factor of n with p_i = 1 mod 4. - Orson R. L. Peters, Jan 31 2017
a(n) = [x^(n^2)] theta_3(x)^2, where theta_3() is the Jacobi theta function. - Ilya Gutkovskiy, Apr 20 2018
From Hugo Pfoertner, Sep 21 2023: (Start)
a(n) = 8*A063014(n) - 4 for n > 0.
a(n) = 4*A256452(n) for n > 0. (End)
EXAMPLE
a(5) = 12 because the circumference of the circle with radius 5 will pass through the twelve points (5, 0), (4, 3), (3, 4), (0, 5), (-3, 4), (-4, 3), (-5, 0), (-4, -3), (-3, -4), (0, -5), (3, -4) and (4, -3). Alternatively, we can say the twelve Gaussian integers 5, 4 + 3i, ... , 4 - 3i all have absolute value of 5.
MAPLE
N:= 1000: # to get a(0) to a(N)
A:= Array(0..N):
A[0]:= 1:
for x from 1 to N do
A[x]:= A[x]+4;
for y from 1 to min(x-1, floor(sqrt(N^2-x^2))) do
z:= x^2+y^2;
if issqr(z) then
t:= sqrt(z);
A[t]:= A[t]+8;
fi
od
od:
seq(A[i], i=0..N); # Robert Israel, May 08 2015
MATHEMATICA
Table[Length[Select[Flatten[Table[r + I i, {r, -n, n}, {i, -n, n}]], Abs[#] == n &]], {n, 0, 49}] (* Alonso del Arte, Feb 11 2012 *)
PROG
(Haskell)
a046109 n = length [(x, y) | x <- [-n..n], y <- [-n..n], x^2 + y^2 == n^2]
-- Reinhard Zumkeller, Jan 23 2012
(Python)
from sympy import factorint
def a(n):
r = 1
for p, e in factorint(n).items():
if p%4 == 1: r *= 2*e + 1
return 4*r if n > 0 else 0
# Orson R. L. Peters, Jan 31 2017
(PARI) a(n)=if(n==0, return(1)); my(f=factor(n)); 4*prod(i=1, #f~, if(f[i, 1]%4==1, 2*f[i, 2]+1, 1)) \\ Charles R Greathouse IV, Feb 01 2017
(PARI) a(n)=if(n==0, return(1)); t=0; for(x=1, n-1, y=n^2-x^2; if(issquare(y), t++)); return(4*t+4) \\ Arkadiusz Wesolowski, Nov 14 2017
KEYWORD
nonn,easy,nice,changed
STATUS
approved