OFFSET
1,2
COMMENTS
Number of images of the map (x,y) -> x^2+y^2 in Z_n.
Let n = p^e and k = r*p^b (gcd(r,p) = 1). If p == 1 (mod 4), then x^2 + y^2 == k (mod p) always have solutions; if p == 3 (mod 4), then x^2 + y^2 == k (mod p) is solvable if and only if b is even or b >= e; if p = 2, then x^2 + y^2 == k (mod p) is solvable if and only if r == 1 (mod 4) or b >= e - 1. If 0 <= k < n, then the number of solutions to x^2 + y^2 == k (mod n) is A305191(n,k). - Jianing Song, Apr 20 2019
LINKS
Jianing Song, Table of n, a(n) for n = 1..10000 (first 1000 terms from Michel Marcus)
FORMULA
Multiplicative with a(p^e) = p^e if p == 1 (mod 4); ceiling(p^(e+1)/(p+1)) if p == 3 (mod 4); 2^(e-1) + 1 if p = 2. - Jianing Song, Apr 20 2019
Sum_{k=1..n} a(k) ~ c * n^2, where c = (11/24) * Product_{p prime == 3 (mod 4)} (1 - 1/p^3)/(1 - 1/p^4) = (11/24) * A334427/A334448 = 0.44532386516028771931... . - Amiram Eldar, Feb 17 2024
MATHEMATICA
(For[v = Table[0, {m, 1, n^2}]; m = 1; i = 0, i < n, i++, For[j = 0, j < n, j++, v[[m]] = Mod[i^2 + j^2, n]; m = m + 1]]; Length[Union[v]])
(* Second program: *)
a[n_] := Module[{p, e}, Product[{p, e} = pe; Which[Mod[p, 4] == 1, p^e, Mod[p, 4] == 3, Ceiling[p^(e+1)/(p+1)], p == 2, 2^(e-1) + 1, True, p], {pe, FactorInteger[n]}]];
Array[a, 100] (* Jean-François Alcover, Jul 30 2020 *)
PROG
(PARI) a(n) = #Set(vector(n^2, i, ((i%n)^2 + (i\n)^2) % n)); \\ Michel Marcus, Jul 08 2017
(PARI) a(n)=
{
my(r=1, f=factor(n));
for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]);
if(p==2, r*=2^(e-1)+1);
if(p%4==1, r*=p^e);
if(p%4==3, r*=ceil(p^(e+1)/(p+1)));
);
return(r);
} \\ Jianing Song, Apr 20 2019
CROSSREFS
KEYWORD
mult,nonn
AUTHOR
Steven Finch, Jan 30 2009
STATUS
approved