OFFSET
1,2
COMMENTS
For any n > 2, there are quadratic nonresidues mod n, so a(n) < n. - Charles R Greathouse IV, Oct 28 2022
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000
Imanuel Chen and Michael Z. Spivey, Integral Generalized Binomial Coefficients of Multiplicative Functions, Preprint 2015; Summer Research Paper 238, Univ. Puget.
Steven R. Finch and Pascal Sebah, Squares and Cubes Modulo n, arXiv:math/0604465 [math.NT], 2006-2016.
Shuguang Li, On the number of elements with maximal order in the multiplicative group modulo n, Acta Arithm. 86 (2) (1998) 113, see proof of theorem 2.1.
Param Parekh, Paavan Parekh, Sourav Deb, and Manish K. Gupta, On the Classification of Weierstrass Elliptic Curves over Z_n, arXiv:2310.11768 [cs.CR], 2023. See p. 6.
E. J. F. Primrose, The number of quadratic residues mod m, Math. Gaz. v. 61 (1977) n. 415, 60-61.
Walter D. Stangl, Counting Squares in Z_n, Math. Mag. 69 (1996) 285-289.
FORMULA
a(n) = A105612(n) + 1.
Multiplicative with a(p^e) = floor(p^e/6) + 2 if p = 2; floor(p^(e+1)/(2p + 2)) + 1 if p > 2. - David W. Wilson, Aug 01 2001
a(2^n) = A023105(n). a(3^n) = A039300(n). a(5^n) = A039302(n). a(7^n) = A039304(n). - R. J. Mathar, Sep 28 2017
Sum_{k=1..n} a(k) ~ c * n^2/sqrt(log(n)), where c = (17/(32*sqrt(Pi))) * Product_{p prime} (1 - (p^2+2)/(2*(p^2+1)*(p+1))) * (1-1/p)^(-1/2) = 0.37672933209687137604... (Finch and Sebah, 2006). - Amiram Eldar, Oct 18 2022
EXAMPLE
The sequence of squares (A000290) modulo 10 reads 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1,... and this reduced sequence contains a(10) = 6 different values, {0,1,4,5,6,9}. - R. J. Mathar, Oct 10 2014
MAPLE
A000224 := proc(m)
{seq( modp(b^2, m), b=0..m-1) };
nops(%) ;
end proc: # Emeric Deutsch
# 2nd implementation
A000224 := proc(n)
local a, ifs, f, p, e, c ;
a := 1 ;
ifs := ifactors(n)[2] ;
for f in ifs do
p := op(1, f) ;
e := op(2, f) ;
if p = 2 then
if type(e, 'odd') then
a := a*(2^(e-1)+5)/3 ;
else
a := a*(2^(e-1)+4)/3 ;
end if;
else
if type(e, 'odd') then
c := 2*p+1 ;
else
c := p+2 ;
end if;
a := a*(p^(e+1)+c)/2/(p+1) ;
end if;
end do:
a ;
end proc: # R. J. Mathar, Oct 10 2014
MATHEMATICA
Length[Union[#]]& /@ Table[Mod[k^2, n], {n, 65}, {k, n}] (* Jean-François Alcover, Aug 30 2011 *)
a[2] = 2; a[n_] := a[n] = Switch[fi = FactorInteger[n], {{_, 1}}, (fi[[1, 1]] + 1)/2, {{2, _}}, 3/2 + 2^fi[[1, 2]]/6 + (-1)^(fi[[1, 2]]+1)/6, {{_, _}}, {p, k} = fi[[1]]; 3/4 + (p-1)*(-1)^(k+1)/(4*(p+1)) + p^(k+1)/(2*(p+1)), _, Times @@ Table[ a[Power @@ f], {f, fi}]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Mar 09 2015 *)
PROG
(PARI) a(n) = local(v, i); v = vector(n, i, 0); for(i=0, floor(n/2), v[i^2%n+1] = 1); sum(i=1, n, v[i]) \\ Franklin T. Adams-Watters, Nov 05 2006
(PARI) a(n)=my(f=factor(n)); prod(i=1, #f[, 1], if(f[i, 1]==2, 2^f[1, 2]\6+2, f[i, 1]^(f[i, 2]+1)\(2*f[i, 1]+2)+1)) \\ Charles R Greathouse IV, Jul 15 2011
(Haskell)
a000224 n = product $ zipWith f (a027748_row n) (a124010_row n) where
f 2 e = 2 ^ e `div` 6 + 2
f p e = p ^ (e + 1) `div` (2 * p + 2) + 1
-- Reinhard Zumkeller, Aug 01 2012
(Python)
from math import prod
from sympy import factorint
def A000224(n): return prod((p**(e+1)//((p+1)*(q:=1+(p==2)))>>1)+q for p, e in factorint(n).items()) # Chai Wah Wu, Oct 07 2024
CROSSREFS
KEYWORD
nonn,easy,nice,mult
AUTHOR
STATUS
approved