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 A046530 Number of distinct cubic residues mod n. 25
 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 S. R. Finch and Pascal Sebah, Squares and Cubes Modulo n, arXiv:math/0604465 [math.NT], 2005-2016. S. 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 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

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Last modified October 17 02:12 EDT 2018. Contains 316275 sequences. (Running on oeis4.)