cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A046530 Number of distinct cubic residues mod n.

Original entry on oeis.org

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
Offset: 1

Views

Author

David W. Wilson

Keywords

Comments

In other words, number of distinct cubes mod n. - N. J. A. Sloane, Oct 05 2024
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

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.

Programs

  • Haskell
    import Data.List (nub)
a046530 n = length $ nub $ map (`mod` n) $
                           take (fromInteger n) $ tail a000578_list
-- Reinhard Zumkeller, Aug 01 2012
  • 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 *)
  • 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*iAndrew Howroyd, Jul 17 2018

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

A257301 Number of cubic nonresidues modulo n.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 4, 3, 6, 0, 0, 3, 8, 8, 0, 6, 0, 12, 12, 5, 12, 0, 0, 9, 4, 16, 20, 19, 0, 0, 20, 13, 0, 0, 20, 27, 24, 24, 24, 15, 0, 24, 28, 11, 30, 0, 0, 18, 34, 8, 0, 37, 0, 40, 0, 41, 36, 0, 0, 15, 40, 40, 54, 27, 40, 0, 44, 17, 0, 40, 0, 57, 48, 48, 12, 55, 44, 48, 52, 30
Offset: 1

Views

Author

Stanislav Sykora, Apr 19 2015

Keywords

Comments

a(n) is the number of values r, 0<=r=0, (m^p)%n != r. Compared to quadratic nonresidues (p=2, sequence A095972), the most evident difference is the frequent occurrence of a(n)=0 (for values of n which belong to A074243).

Examples

			a(5)=0, because the set {(k^3)%5}, with k=0..4, evaluates to {0,1,3,2,4},
        with no missing residue values.
a(7)=4, because the set {(k^3)%7}, with k=0..6, evaluates to
        {0,1,1,6,1,6,6}, with missing residue values {2,3,4,5}.

Links

Crossrefs

Nonresidues for other exponents: A095972 (p=2), A257302 (p=4), A257303 (p=5).
Cf. A074243, A046530.

Programs

  • Maple
    seq(n - nops({seq(a^3 mod n,a=0..n-1)}), n=1..100); # Robert Israel, Apr 20 2015
  • Mathematica
    Table[Length[Complement[Range[n - 1], Union[Mod[Range[n]^3, n]]]], {n, 100}] (* Vincenzo Librandi, Apr 20 2015 *)
  • PARI
    nrespowp(n,p) = {my(v=vector(n),d=0);
  for(r=0,n-1,v[1+(r^p)%n]+=1);
  for(k=1,n,if(v[k]==0,d++));
  return(d);}
a(n) = nrespowp(n,3)
  • 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)); n-prod(i=1, #f~, g(f[i,1], f[i,2])) \\ Charles R Greathouse IV, Apr 20 2015

Formula

a(n) = n - A046530(n).
Satisfies a(A074243(n))=0.
Satisfies a(n) <= n-3 (residues 0, 1, and n-1 are always present).
a(n) = n - A046530(n). - Robert Israel, Apr 20 2015

A276920 Number of solutions to x^3 + y^3 + z^3 + t^3 == 0 (mod n) for 1 <= x, y, z, t <= n.

Original entry on oeis.org

1, 8, 27, 72, 125, 216, 595, 704, 1539, 1000, 1331, 1944, 3133, 4760, 3375, 5632, 4913, 12312, 8911, 9000, 16065, 10648, 12167, 19008, 16125, 25064, 45927, 42840, 24389, 27000, 35371, 47104, 35937, 39304, 74375, 110808, 58645, 71288, 84591, 88000
Offset: 1

Views

Author

José María Grau Ribas, Sep 22 2016

Keywords

Comments

a(n) = n^3 if n is in A074243. - Robert Israel, Oct 13 2016

Examples

			For n = 3, we see that all nondecreasing solutions {x, y, z, t} are in {{1, 1, 1, 3}, {1, 1, 2, 2}, {1, 2, 3, 3}, {2, 2, 2, 3}, {3, 3, 3, 3}}. The numbers in the sets can be ordered in 4, 6, 12, 4 and 1 ways respectively. Therefore, a(3) = 4 + 6 + 12 + 4 + 1 = 27. - _David A. Corneth_, Oct 11 2016

Links

Crossrefs

Cf. A000189, A047726, A060839, A063454, A074243, A087412, A087786, A254073, A276919.

Programs

  • Maple
    CF:= table([[false, false, true] = 12, [true, false, false] = 12, [true, false, true] = 6, [false, false, false] = 24, [true, true, true] = 1, [false, true, true] = 4, [false, true, false] = 12, [true, true, false] = 4]):
f1:= proc(n)
  option remember;
  local count, t, x,y,z,signature;
  if isprime(n) and n mod 3 = 2 then return n^3 fi;
  count:= 0;
  for t from 1 to n do
    for x from 1 to t do
      for y from 1 to x do
        for z from 1 to y do
          if t^3 + x^3 + y^3 + z^3 mod n = 0 then
            signature:= map(evalb,[z=y,y=x,x=t]);
            count:= count + CF[signature];
          fi
  od od od od;
  count
end proc:
f:= proc(n) local t;
    mul(f1(t[1]^t[2]),t=ifactors(n)[2])
end proc:
map(f, [$1..40]); # Robert Israel, Oct 13 2016
  • Mathematica
    JJJ[4, n, lam] = Sum[If[Mod[a^3 + b^3 + c^3 + d^3, n] == Mod[lam, n], 1, 0], {d, 0, n - 1}, {a, 0, n - 1}, {b, 0, n - 1}, {c, 0 , n - 1}]; Table[JJJ[4, n, 0], {n, 1, 50}]
  • PARI
    a(n) = sum(x=1, n, sum(y=1, n, sum(z=1, n, sum(t=1, n, Mod(x,n)^3 + Mod(y,n)^3 + Mod(z,n)^3 + Mod(t,n)^3 == 0)))); \\ Michel Marcus, Oct 11 2016
  • PARI
    qperms(v) = {my(r=1,t); v = vecsort(v); for(i=1,#v-1, if(v[i]==v[i+1], t++, r*=binomial(i, t+1);t=0));r*=binomial(#v,t+1)}
a(n) = {my(t=0); forvec(v=vector(4,i,[1,n]), if(sum(i=1,4,Mod(v[i],n)^3)==0, t+=qperms(v)),1);t} \\ David A. Corneth, Oct 11 2016
  • Python
    def A276920(n):
    ndict = {}
    for i in range(n):
        i3 = pow(i,3,n)
        for j in range(i+1):
            j3 = pow(j,3,n)
            m = (i3+j3) % n
            if m in ndict:
                if i == j:
                    ndict[m] += 1
                else:
                    ndict[m] += 2
            else:
                if i == j:
                    ndict[m] = 1
                else:
                    ndict[m] = 2
    count = 0
    for i in ndict:
        j = (-i) % n
        if j in ndict:
            count += ndict[i]*ndict[j]
    return count # Chai Wah Wu, Jun 06 2017
Showing 1-3 of 3 results.

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