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 1-9 of 9 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

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Author

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

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).

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

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).

A365099 Number of distinct residues of x^n (mod n^2), x=0..n^2-1.

Original entry on oeis.org

1, 2, 3, 2, 5, 4, 7, 3, 7, 6, 11, 4, 13, 8, 15, 5, 17, 8, 19, 4, 9, 12, 23, 6, 21, 14, 19, 8, 29, 12, 31, 9, 33, 18, 35, 8, 37, 20, 15, 6, 41, 8, 43, 12, 35, 24, 47, 10, 43, 22, 51, 8, 53, 20, 15, 12, 21, 30, 59, 8, 61, 32, 21, 17, 65, 24, 67, 10, 69, 24, 71, 12, 73, 38, 63
Offset: 1

Views

Author

Albert Mukovskiy, Aug 21 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = #Set(vector(n^2, x, Mod(x-1,n^2)^n)); \\ Michel Marcus, Aug 22 2023
    
  • Python
    def A365099(n): return len({pow(x,n,n**2) for x in range(n**2)}) # Chai Wah Wu, Aug 22 2023

A365104 Number of distinct quintic residues x^5 (mod 5^n), x=0..5^n-1.

Original entry on oeis.org

1, 5, 5, 21, 101, 501, 2505, 12505, 62521, 312601, 1563001, 7815005, 39075005, 195375021, 976875101, 4884375501, 24421877505, 122109387505, 610546937521, 3052734687601, 15263673438001, 76318367190005, 381591835950005, 1907959179750021, 9539795898750101, 47698979493750501, 238494897468752505, 1192474487343762505, 5962372436718812521, 29811862183594062601
Offset: 0

Views

Author

Albert Mukovskiy, Aug 24 2023

Keywords

Comments

It appears that for a prime p>2 the number of distinct residues x^p (mod p^n) is a(n) = (p-1)*p^(n-2) + a(n-p), with a(n<1)=1, a(1)=p.

Crossrefs

Programs

  • Mathematica
    a[n_]:=CountDistinct[Table[PowerMod[x-1, 5, 5^(n-1)], {x, 1, 5^(n-1)}]]; Array[a, 13]
  • Python
    def A365104(n): return len({pow(x,5,5**n) for x in range(5**n)}) # Chai Wah Wu, Sep 17 2023

Formula

For n >= 6, a(n) = 4*5^(n-2) + a(n-5) = 5*a(n-1) + a(n-5) - 5*a(n-6). O.g.f: (-5*x^5 - 4*x^4 - 4*x^3 - 20*x^2 + 1)/(5*x^6 - x^5 - 5*x + 1). - Max Alekseyev, Feb 19 2024

Extensions

Terms a(16) onward from Max Alekseyev, Feb 19 2024

A367484 Number of integers of the form (x^4 + y^4) mod 3^n; a(n) = A289559(3^n).

Original entry on oeis.org

1, 3, 7, 19, 55, 165, 493, 1477, 4429, 13287, 39859, 119575, 358723, 1076169, 3228505, 9685513, 29056537
Offset: 0

Views

Author

Albert Mukovskiy, Nov 19 2023

Keywords

Comments

It appears that for n > 4: a(n) = 2*3^(n-1) + a(n-4).
For n < 5: a(n) = 2*3^(n-1) + 1.
Conjecture in closed form: a(n) = 2*ceiling(3^(n+3)/80) - 1.

Crossrefs

Subsequence of A289559.

Programs

  • PARI
    a(n) = #setbinop((x, y)->Mod(x,3^n)^4+Mod(y,3^n)^4, [0..3^n-1]);
    
  • Python
    def A367484(n):
        m = 3**n
        return len({(pow(x,4,m)+pow(y,4,m))%m for x in range(m) for y in range(x+1)}) # Chai Wah Wu, Jan 23 2024

Formula

Conjecture: a(n) = 2*ceiling(3^(n+3)/80) - 1.
a(n) = A289559(3^n). - Thomas Scheuerle, Nov 20 2023

A046634 Number of cubic residues mod 6^n.

Original entry on oeis.org

1, 6, 9, 35, 210, 1083, 6253, 37518, 222705, 1331099, 7986594, 47871651, 287102581, 1722615486, 10334532969, 62003849075, 372023094450, 2232108315723, 13392560190013, 80355361140078, 482131358602785
Offset: 0

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Author

Keywords

Programs

  • Mathematica
    LinearRecurrence[{6,0,36,-216,0,-251,1506,0,216,-1296},{1,6,9,35,210,1083,6253,37518,222705,1331099},30] (* Harvey P. Dale, Mar 17 2023 *)

Formula

From R. J. Mathar, Feb 27 2011: (Start)
a(n) = A046530(6^n) = A046631(n)*A046630(n).
a(n) = +6*a(n-1) +36*a(n-3) -216*a(n-4) -251*a(n-6) +1506*a(n-7) +216*a(n-9) -1296*a(n-10).
G.f.: ( 1-27*x^2-55*x^3+795*x^5+690*x^6-2808*x^8-1296*x^9 ) / ( (x-1) *(6*x-1) *(3*x-1) *(2*x-1) *(1+x+x^2) *(4*x^2+2*x+1) *(9*x^2+3*x+1) ). (End)

A046637 Number of cubic residues mod 9^n.

Original entry on oeis.org

1, 3, 21, 169, 1515, 13629, 122641, 1103763, 9933861, 89404729, 804642555, 7241782989, 65176046881, 586584421923, 5279259797301, 47513338175689, 427620043581195, 3848580392230749, 34637223530076721, 311735011770690483
Offset: 0

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Author

Keywords

Programs

  • Mathematica
    LinearRecurrence[{9,0,1,-9},{1,3,21,169},30] (* Harvey P. Dale, Oct 09 2017 *)

Formula

a(n) = A046530(9^n) = A046631(2n). G.f. ( 1-6*x-6*x^2-21*x^3 ) / ( (x-1)*(9*x-1)*(1+x+x^2) ). - R. J. Mathar, Feb 28 2011
If n>=1, a(n) = 9*a(n-1) -6 if n is not a multiple of 3, otherwise a(n) = 9*a(n-1) -20. - Vincenzo Librandi, Mar 18 2011
Showing 1-9 of 9 results.