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.

A063938 Numbers k that divide tau(k), where tau(k)=A000594(k) is Ramanujan's tau function.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, 28, 30, 32, 35, 36, 40, 42, 45, 48, 49, 50, 54, 56, 60, 63, 64, 70, 72, 75, 80, 81, 84, 88, 90, 91, 92, 96, 98, 100, 105, 108, 112, 115, 120, 125, 126, 128, 135, 140, 144, 147, 150, 160, 161, 162, 168
Offset: 1

Views

Author

Robert G. Wilson v, Aug 31 2001

Keywords

Comments

Although most small numbers are in the sequence, it becomes sparser for larger values; e.g., only 504 numbers up to 10000 and only 184 numbers from 10001 to 20000 are in the sequence.

Crossrefs

For the sequence when n is prime see A007659.

Programs

  • Mathematica
    (* First do <Michael De Vlieger, Dec 23 2017 *)
  • PARI
    for (n=1,1000,if(Mod(ramanujantau(n),n)==0,print1(n", "))) \\ Dana Jacobsen, Sep 06 2015
    
  • Perl
    use ntheory ":all"; my @p = grep { !(ramanujan_tau($) % $) } 1..1000; say "@p"; # Dana Jacobsen, Sep 06 2015
    
  • Python
    from itertools import count, islice
    from sympy import divisor_sigma
    def A063938_gen(startvalue=1): # generator of terms >= startvalue
        return filter(lambda n: not -840*(pow(m:=n+1>>1,2,n)*(0 if n&1 else pow(m*divisor_sigma(m),2,n))+(sum(pow(i,4,n)*divisor_sigma(i)*divisor_sigma(n-i) for i in range(1,m))<<1)) % n, count(max(startvalue,1)))
    A063938_list = list(islice(A063938_gen(),25)) # Chai Wah Wu, Nov 08 2022

Extensions

More terms from Dean Hickerson, Jan 03 2003

A296991 Numbers k such that k^2 divides tau(k), where tau(k) = A000594(k) is Ramanujan's tau function.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 21, 24, 27, 32, 36, 40, 42, 48, 54, 64, 72, 81, 84, 96, 108, 120, 128, 135, 144, 162, 168, 189, 192, 216, 243, 256, 270, 280, 288, 324, 336, 360, 378, 384, 432, 448, 486, 512, 540, 576, 640, 648, 672, 729, 756, 768, 828, 840, 864
Offset: 1

Views

Author

Seiichi Manyama, Dec 22 2017

Keywords

Comments

2^k is a term for k >= 0.

Crossrefs

Programs

  • Mathematica
    fQ[n_] := Mod[RamanujanTau@n, n^2] == 0; Select[Range@875, fQ] (* Robert G. Wilson v, Dec 23 2017 *)
  • PARI
    is(n) = Mod(ramanujantau(n), n^2)==0 \\ Felix Fröhlich, Dec 24 2017
    
  • Python
    from itertools import count, islice
    from sympy import divisor_sigma
    def A296991_gen(startvalue=1): # generator of terms >= startvalue
        return filter(lambda n: not -24*((m:=n+1>>1)**2*(0 if n&1 else m*(35*m - 52*n)*divisor_sigma(m)**2)+sum(i**3*(70*i - 140*n)*divisor_sigma(i)*divisor_sigma(n-i) for i in range(1,m))) % n**2, count(max(startvalue,1)))
    A296991_list = list(islice(A296991_gen(),20)) # Chai Wah Wu, Nov 08 2022

A296992 Largest number m such that n^m divides tau(n), where tau(n) = A000594(n) is Ramanujan's tau function.

Original entry on oeis.org

3, 2, 3, 1, 3, 1, 3, 2, 1, 0, 2, 0, 1, 1, 3, 0, 2, 0, 1, 2, 0, 0, 3, 1, 0, 2, 1, 0, 1, 0, 3, 0, 0, 1, 2, 0, 0, 0, 2, 0, 2, 0, 0, 1, 0, 0, 2, 1, 1, 0, 0, 0, 2, 0, 1, 0, 0, 0, 1, 0, 0, 1, 3, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 1, 0, 0, 0, 0, 1, 2, 0, 0, 2, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0
Offset: 2

Views

Author

Seiichi Manyama, Dec 22 2017

Keywords

Examples

			tau(2) =   -24 and 2^3 divides   24, so a(2) = 3.
tau(3) =   252 and 3^2 divides  252, so a(3) = 2.
tau(4) = -1472 and 4^3 divides 1472, so a(4) = 3.
		

Crossrefs

Cf. A063938 (a(n)>=1), A296991 (a(n)>=2), A296993 (a(n)>=3).
Cf. A191599 (a(n)=0), A297000 (a(n)=1), A297001 (a(n)=2).

Programs

  • Mathematica
    f[n_] := Block[{m = 0}, While[Mod[RamanujanTau@n, n^m] == 0, m++]; m - 1]; Array[f, 93, 2] (* Robert G. Wilson v, Dec 23 2017 *)
    a[n_] := IntegerExponent[RamanujanTau[n], n]; Array[a, 100, 2] (* Amiram Eldar, Jan 09 2025 *)
  • PARI
    a(n) = valuation(ramanujantau(n), n); \\ Amiram Eldar, Jan 09 2025
Showing 1-3 of 3 results.