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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

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