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A000224 Number of squares mod n. 53
1, 2, 2, 2, 3, 4, 4, 3, 4, 6, 6, 4, 7, 8, 6, 4, 9, 8, 10, 6, 8, 12, 12, 6, 11, 14, 11, 8, 15, 12, 16, 7, 12, 18, 12, 8, 19, 20, 14, 9, 21, 16, 22, 12, 12, 24, 24, 8, 22, 22, 18, 14, 27, 22, 18, 12, 20, 30, 30, 12, 31, 32, 16, 12, 21, 24, 34, 18, 24, 24, 36, 12 (list; graph; refs; listen; history; text; internal format)
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
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
CROSSREFS
Cf. A095972, A046530 (cubic residues), A052273 (4th powers), A052274 (5th powers), A052275 (6th powers), A085310 (7th powers), A085311 (8th powers), A085312 (9th powers), A085313 (10th powers), A085314 (11th powers), A228849 (12th powers).
Sequence in context: A318816 A085202 A096009 * A085201 A300401 A051601
KEYWORD
nonn,easy,nice,mult
AUTHOR
STATUS
approved

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Last modified March 28 14:21 EDT 2024. Contains 371254 sequences. (Running on oeis4.)