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A046530 Number of distinct cubic residues mod n. 27
1, 2, 3, 3, 5, 6, 3, 5, 3, 10, 11, 9, 5, 6, 15, 10, 17, 6, 7, 15, 9, 22, 23, 15, 21, 10, 7, 9, 29, 30, 11, 19, 33, 34, 15, 9, 13, 14, 15, 25, 41, 18, 15, 33, 15, 46, 47, 30, 15, 42, 51, 15, 53, 14, 55, 15, 21, 58, 59, 45, 21, 22, 9, 37, 25, 66, 23, 51, 69, 30, 71, 15, 25, 26, 63 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Cubic analog of A000224. - Steven Finch, Mar 01 2006

A074243 contains values of n such that a(n) = n. - Dmitri Kamenetsky, Nov 03 2012

LINKS

T. D. Noe, Table of n, a(n) for n = 1..1000

Steven R. Finch and Pascal Sebah, Squares and Cubes Modulo n, arXiv:math/0604465 [math.NT], 2005-2016.

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.

FORMULA

a(n) = n - A257301(n). - Stanislav Sykora, Apr 21 2015

a(2^n) = A046630(n). a(3^n) = A046631(n). a(5^n) = A046633(n). a(7^n) = A046635(n). - R. J. Mathar, Sep 28 2017

Multiplicative with a(p^e) = 1 + Sum_{i=0..floor((e-1)/3)} (p - 1)*p^(e-3*i-1)/k where k = 3 if (p = 3 and 3*i + 1 = e) or (p mod 3 = 1) otherwise k = 1. - Andrew Howroyd, Jul 17 2018

Sum_{k=1..n} a(k) ~ c * n^2/log(n)^(1/3), where c = (6/(13*Gamma(2/3))) * (2/3)^(-1/3) * Product_{p prime == 2 (mod 3)} (1 - (p^2+1)/((p^2+p+1)*(p^2-p+1)*(p+1))) * (1-1/p)^(-1/3) * Product_{p prime == 1 (mod 3)} (1 - (2*p^4+3*p^2+3)/(3*(p^2+p+1)*(p^2-p+1)*(p+1))) * (1-1/p)^(-1/3) = 0.48487418844474389597... (Finch and Sebah, 2006). - Amiram Eldar, Oct 18 2022

MAPLE

A046530 := proc(n)

local a, pf ;

a := 1 ;

if n = 1 then

return 1;

end if;

for i in ifactors(n)[2] do

p := op(1, i) ;

e := op(2, i) ;

if p = 3 then

if e mod 3 = 0 then

a := a*(3^(e+1)+10)/13 ;

elif e mod 3 = 1 then

a := a*(3^(e+1)+30)/13 ;

else

a := a*(3^(e+1)+12)/13 ;

end if;

elif p mod 3 = 2 then

if e mod 3 = 0 then

a := a*(p^(e+2)+p+1)/(p^2+p+1) ;

elif e mod 3 = 1 then

a := a*(p^(e+2)+p^2+p)/(p^2+p+1) ;

else

a := a*(p^(e+2)+p^2+1)/(p^2+p+1) ;

end if;

else

if e mod 3 = 0 then

a := a*(p^(e+2)+2*p^2+3*p+3)/3/(p^2+p+1) ;

elif e mod 3 = 1 then

a := a*(p^(e+2)+3*p^2+3*p+2)/3/(p^2+p+1) ;

else

a := a*(p^(e+2)+3*p^2+2*p+3)/3/(p^2+p+1) ;

end if;

end if;

end do:

a ;

end proc:

seq(A046530(n), n=1..40) ; # R. J. Mathar, Nov 01 2011

MATHEMATICA

Length[Union[#]]& /@ Table[Mod[k^3, n], {n, 75}, {k, n}] (* Jean-François Alcover, Aug 30 2011 *)

Length[Union[#]]&/@Table[PowerMod[k, 3, n], {n, 80}, {k, n}] (* Harvey P. Dale, Aug 12 2015 *)

PROG

(Haskell)

import Data.List (nub)

a046530 n = length $ nub $ map (`mod` n) $

take (fromInteger n) $ tail a000578_list

-- Reinhard Zumkeller, Aug 01 2012

(PARI) g(p, e)=if(p==3, (3^(e+1)+if(e%3==1, 30, if(e%3, 12, 10)))/13, if(p%3==2, (p^(e+2)+if(e%3==1, p^2+p, if(e%3, p^2+1, p+1)))/(p^2+p+1), (p^(e+2)+if(e%3==1, 3*p^2+3*p+2, if(e%3, 3*p^2+2*p+3, 2*p^2+3*p+3)))/3/(p^2+p+1)))

a(n)=my(f=factor(n)); prod(i=1, #f[, 1], g(f[i, 1], f[i, 2])) \\ Charles R Greathouse IV, Jan 03 2013

(PARI) a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); 1 + sum(i=0, (e-1)\3, if(p%3==1 || (p==3&&3*i<e-1), 1/3, 1)*(p-1)*p^(e-3*i-1)) )} \\ Andrew Howroyd, Jul 17 2018

CROSSREFS

For number of k-th power residues mod n, see A000224 (k=2), A052273 (k=4), A052274 (k=5), A052275 (k=6), A085310 (k=7), A085311 (k=8), A085312 (k=9), A085313 (k=10), A085314 (k=12), A228849 (k=13).

Cf. A000578, A087786, A257301.

Sequence in context: A289630 A023160 A085312 * A003558 A216066 A234094

Adjacent sequences: A046527 A046528 A046529 * A046531 A046532 A046533

KEYWORD

nonn,mult,easy,nice

AUTHOR

David W. Wilson

STATUS

approved

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Last modified December 5 19:04 EST 2022. Contains 358588 sequences. (Running on oeis4.)