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 1-4 of 4 results.

A293482 The number of 5th powers in the multiplicative group modulo n.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 2, 4, 12, 6, 8, 8, 16, 6, 18, 8, 12, 2, 22, 8, 4, 12, 18, 12, 28, 8, 6, 16, 4, 16, 24, 12, 36, 18, 24, 16, 8, 12, 42, 4, 24, 22, 46, 16, 42, 4, 32, 24, 52, 18, 8, 24, 36, 28, 58, 16, 12, 6, 36, 32, 48, 4, 66, 32, 44, 24, 14, 24, 72, 36, 8, 36, 12, 24, 78, 32, 54, 8, 82, 24
Offset: 1

Views

Author

R. J. Mathar, Oct 10 2017

Keywords

Comments

The size of the set of numbers j^5 mod n, gcd(j,n)=1, 1 <= j <= n.
A000010(n) / a(n) is another multiplicative integer sequence.

Links

Crossrefs

The number of k-th powers in the multiplicative group modulo n: A046073 (k=2), A087692 (k=3), A250207 (k=4), this sequence (k=5), A293483 (k=6), A293484 (k=7), A293485 (k=8).
Cf. A052274, A319099, A000010.

Programs

  • Maple
    A293482 := proc(n)
    local r,j;
    r := {} ;
    for j from 1 to n do
        if igcd(j,n)= 1 then
            r := r union { modp(j &^ 5,n) } ;
        end if;
    end do:
    nops(r) ;
end proc:
seq(A293482(n),n=1..120) ;
  • Mathematica
    a[n_] := Module[{r, j}, r = {}; For[j = 1, j <= n, j++, If[GCD[j, n] == 1, r = r ~Union~ {PowerMod[j, 5, n]}] ]; Length[r]];
Table[a[n], {n, 1, 120}] (* Jean-François Alcover, Feb 14 2023, after R. J. Mathar *)
f[p_, e_] := (p - 1)*p^(e - 1)/If[Mod[p, 5] == 1, 5, 1]; f[2, e_] := 2^(e - 1); f[2, 1] = 1; f[5, e_] := 4*5^(e-2); f[5, 1] = 4; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Aug 10 2023 *)

Formula

Conjecture: a(2^e) = 1 for e <= 1; a(2^e) = 2^(e-1) for e >= 1; a(5)=4; a(5^e) = 4*5^(e-2) for e > 1; a(p^e) = (p-1)*p^(e-1) for p == {2,3,4} (mod 5); a(p^e) = (p-1)*p^(e-1)/5 for p == 1 (mod 5). - R. J. Mathar, Oct 13 2017
a(n) = A000010(n)/A319099(n). This implies that the conjecture above is true. - Jianing Song, Nov 10 2019

A250207 The number of quartic terms in the multiplicative group modulo n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 5, 1, 3, 3, 1, 1, 4, 3, 9, 1, 3, 5, 11, 1, 5, 3, 9, 3, 7, 1, 15, 2, 5, 4, 3, 3, 9, 9, 3, 1, 10, 3, 21, 5, 3, 11, 23, 1, 21, 5, 4, 3, 13, 9, 5, 3, 9, 7, 29, 1, 15, 15, 9, 4, 3, 5, 33, 4, 11, 3, 35, 3, 18, 9, 5, 9, 15, 3, 39, 1
Offset: 1

Views

Author

R. J. Mathar, Mar 02 2015

Keywords

Comments

In the character table of the multiplicative group modulo n there are phi(n) different characters. [This is made explicit for example by the number of rows in arXiv:1008.2547.] The set of the fourth powers of the characters in all representations has some cardinality, which defines the sequence.

Examples

			For n <= 6, the set of all characters in all representations consists of a subset of +1, -1, +i or -i. Their fourth powers are all +1, a single value, so a(n)=1 then.
For n=7, the set of characters is 1, -1, +-1/2 +- sqrt(3)*i/2, so their fourth powers are 1 or -1/2 +- sqrt(3)*i/2, which are three different values, so a(7)=3.
For n=11, the fourth powers of the characters may be 1, exp(+-2*i*Pi/5) or exp(+-4*i*Pi/5), which are 5 different values.

Links

Crossrefs

Cf. A046073, A087692, A052273, A293482 - A293485.

Programs

  • Maple
    A250207 := proc(n)
    numtheory[phi](n)/A073103(n) ;
end proc:
  • Mathematica
    a[n_] := EulerPhi[n]/Count[Range[0, n-1]^4 - 1, k_ /; Divisible[k, n]];
Array[a, 80] (* Jean-François Alcover, Nov 20 2017 *)
f[p_, e_] := (p - 1)*p^(e - 1)/If[Mod[p, 4] == 1, 4, 2]; f[2, e_] := If[e <= 3, 1, 2^(e - 4)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Aug 10 2023 *)
  • PARI
    a(n)=my(f=factor(n)); prod(i=1,#f~, if(f[i,1]==2, 2^max(0,f[i,2]-4), f[i,1]^(f[i,2]-1)*(f[i,1]-1)/if(f[i,1]%4==1,4,2))) \\ Charles R Greathouse IV, Mar 02 2015

Formula

a(n) = A000010(n)/A073103(n).
Multiplicative with a(2^e) = 1 for e<=3; a(2^e) = 2^(e-4) for e>=4; a(p^e) = p^(e-1)*(p-1)/4 for e>=1 and p == 1 (mod 4); a(p^e) = p^(e-1)*(p-1)/2 for e>=1 and p == 3 (mod 4). (Derived from A073103.) - R. J. Mathar, Oct 13 2017

A293483 The number of 6th powers in the multiplicative group modulo n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 5, 1, 2, 1, 2, 2, 8, 1, 3, 2, 1, 5, 11, 1, 10, 2, 3, 1, 14, 2, 5, 4, 5, 8, 2, 1, 6, 3, 2, 2, 20, 1, 7, 5, 2, 11, 23, 2, 7, 10, 8, 2, 26, 3, 10, 1, 3, 14, 29, 2, 10, 5, 1, 8, 4, 5, 11, 8, 11, 2, 35, 1, 12, 6, 10, 3, 5, 2, 13, 4, 9, 20
Offset: 1

Views

Author

R. J. Mathar, Oct 10 2017

Keywords

Comments

The size of the set of numbers j^6 mod n, gcd(j,n)=1, 1 <= j <= n.
A000010(n) / a(n) is another multiplicative integer sequence.

Links

Crossrefs

The number of k-th powers in the multiplicative group modulo n: A046073 (k=2), A087692 (k=3), A250207 (k=4), A293482 (k=5), this sequence (k=6), A293484 (k=7), A293485 (k=8).
Cf. A052275, A319100, A000010.

Programs

  • Maple
    A293483 := proc(n)
    local r,j;
    r := {} ;
    for j from 1 to n do
        if igcd(j,n)= 1 then
            r := r union { modp(j &^ 6,n) } ;
        end if;
    end do:
    nops(r) ;
end proc:
seq(A293483(n),n=1..120) ;
  • Mathematica
    a[n_] := EulerPhi[n]/Count[Range[0, n - 1]^6 - 1, k_ /; Divisible[k, n]];
Array[a, 100] (* Jean-François Alcover, May 24 2023 *)
f[p_, e_] := (p - 1)*p^(e - 1)/If[Mod[p, 6] == 1, 6, 2]; f[2, e_] := If[e <= 3, 1, 2^(e - 3)]; f[3, e_] := If[e <= 2, 1, 3^(e - 2)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Aug 10 2023 *)

Formula

Conjecture: a(2^e) = 1 for e <= 3; a(2^e) = 2^(e-3) for e >= 3; a(3^e) = 1 for e <= 2; a(3^e) = 3^(e-2) for e >= 2; a(p^e) = (p-1)*p^(e-1)/2 for p == 5 (mod 6); a(p^e) = (p-1)*p^(e-1)/6 for p == 1 (mod 6). - R. J. Mathar, Oct 13 2017
a(n) = A000010(n)/A319100(n). This implies that the conjecture above is true. - Jianing Song, Nov 10 2019

A293484 The number of 7th powers in the multiplicative group modulo n.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8, 8, 16, 6, 18, 8, 12, 10, 22, 8, 20, 12, 18, 12, 4, 8, 30, 16, 20, 16, 24, 12, 36, 18, 24, 16, 40, 12, 6, 20, 24, 22, 46, 16, 6, 20, 32, 24, 52, 18, 40, 24, 36, 4, 58, 16, 60, 30, 36, 32, 48, 20, 66, 32, 44, 24, 10, 24, 72, 36, 40, 36
Offset: 1

Views

Author

R. J. Mathar, Oct 10 2017

Keywords

Comments

The size of the set of numbers j^7 mod n, gcd(j,n)=1, 1 <= j <= n.
A000010(n) / a(n) is another multiplicative integer sequence (size of the kernel of the isomorphism of the multiplicative group modulo n to the multiplicative group of 7th powers modulo n).

Links

Crossrefs

The number of k-th powers in the multiplicative group modulo n: A046073 (k=2), A087692 (k=3), A250207 (k=4), A293482 (k=5), A293483 (k=6), this sequence (k=7), A293485 (k=8).
Cf. A085310, A319101, A000010.

Programs

  • Maple
    A293484 := proc(n)
    local r,j;
    r := {} ;
    for j from 1 to n do
        if igcd(j,n)= 1 then
            r := r union { modp(j &^ 7,n) } ;
        end if;
    end do:
    nops(r) ;
end proc:
seq(A293484(n),n=1..120) ;
  • Mathematica
    a[n_] := EulerPhi[n]/Count[Range[0, n - 1]^7 - 1, k_ /; Divisible[k, n]];
Array[a, 100] (* Jean-François Alcover, May 24 2023 *)
f[p_, e_] := (p-1)*p^(e-1)/If[Mod[p, 7] == 1, 7, 1]; f[2, e_] := 2^(e-1); f[7, 1] = 6; f[7, e_] := 6*7^(e-2); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Aug 10 2023 *)

Formula

Conjecture: a(2^e) = 1 for e <= 1; a(2^e) = 2^(e-1) for e >= 1; a(7^e) = 6 for e=1; a(7^e) = 6*7^(e-2) for e >= 2; a(p^e) = (p-1)*p^(e-1) for p == {2,3,4,5,6} (mod 7); a(p^e) = (p-1)*p^(e-1)/7 for p == 1 (mod 7). - R. J. Mathar, Oct 13 2017
a(n) = A000010(n)/A319101(n). This implies that the conjecture above is true. - Jianing Song, Nov 10 2019
Showing 1-4 of 4 results.

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