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.

A365100 Number of distinct residues of x^n (mod n^3), x=0..n^3-1.

Original entry on oeis.org

1, 3, 7, 6, 21, 8, 43, 18, 55, 22, 111, 20, 157, 44, 147, 65, 273, 56, 343, 30, 105, 112, 507, 68, 501, 158, 487, 110, 813, 88, 931, 257, 777, 274, 903, 140, 1333, 344, 371, 102, 1641, 64, 1807, 280, 1155, 508, 2163, 260, 2059, 502, 1911, 200, 2757, 488, 483, 374, 805, 814
Offset: 1

Views

Author

Albert Mukovskiy, Aug 21 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = #Set(vector(n^3, x, Mod(x-1,n^3)^n)); \\ Michel Marcus, Aug 22 2023
    
  • Python
    def A365100(n): return len({pow(x,n,n**3) for x in range(n**3)}) # Chai Wah Wu, Aug 23 2023

A365101 Number of distinct residues of x^n (mod n^4), x=0..n^4-1.

Original entry on oeis.org

1, 4, 21, 18, 101, 30, 295, 130, 487, 153, 1211, 170, 2029, 444, 1919, 1025, 4625, 732, 6499, 442, 1881, 1818, 11639, 1290, 12501, 3045, 13123, 2516, 23549, 1530, 28831, 8193, 23009, 6939, 29795, 4148, 49285, 9750, 12863, 3354, 67241, 1500, 77659, 10302, 49187, 17460, 101615
Offset: 1

Views

Author

Albert Mukovskiy, Aug 21 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = #Set(vector(n^4, x, Mod(x-1,n^4)^n)); \\ Michel Marcus, Aug 22 2023
    
  • Python
    def A365101(n): return len({pow(x,n,n**4) for x in range(n**4)}) # Chai Wah Wu, Aug 23 2023

A365103 Number of distinct quartic residues x^4 (mod 4^n), x=0..4^n-1.

Original entry on oeis.org

1, 2, 2, 6, 18, 70, 274, 1094, 4370, 17478, 69906, 279622, 1118482, 4473926, 17895698, 71582790, 286331154, 1145324614, 4581298450, 18325193798, 73300775186, 293203100742, 1172812402962, 4691249611846, 18764998447378
Offset: 0

Views

Author

Albert Mukovskiy, Aug 24 2023

Keywords

Comments

a(n) = A364811(2n).
For n>=2, A319281(a(n)) == 4^n + [n mod 2 == 1].
For n>=2, a(n)=k: [ A319281(k) == 4^n + [n mod 2 == 1] ].

Crossrefs

Programs

  • Mathematica
    a[n_] = Ceiling[4^n/15] + Boole[Mod[n,2]==1]; Array[a, 24]
  • PARI
    a(n) = ceil(4^n/15)+(Mod(n,2)==1);
    
  • Python
    def A365103(n): return len({pow(x,4,1<<(n<<1)) for x in range(1<<(n<<1))}) # Chai Wah Wu, Sep 18 2023

Formula

a(n) = ceiling(4^n/15) + (n mod 2).

A365102 Number of distinct residues of x^n (mod n^5), x=0..n^5-1.

Original entry on oeis.org

1, 7, 57, 70, 501, 140, 2059, 1029, 4377, 1255, 13311, 1820, 26365, 5150, 27555, 16386, 78609, 10940, 123463, 8190, 37785, 33280, 267675, 28700, 312501, 65915, 354295, 66950, 682893, 35140, 893731, 262145, 732105, 196525, 1031559, 142220, 1823509, 308660
Offset: 1

Views

Author

Albert Mukovskiy, Aug 22 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_]:=CountDistinct[Table[PowerMod[x-1,n,n^5],{x,0,n^5-1}]]; Array[a,38] (* Stefano Spezia, Aug 24 2023 *)
  • PARI
    a(n) = #Set(vector(n^5, x, Mod(x-1, n^5)^n));
    
  • Python
    def A365102(n): return len({pow(x,n,n**5) for x in range(n**5)}) # Chai Wah Wu, Aug 23 2023
Showing 1-4 of 4 results.