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.

A104524 Number of distinct prime divisors of 55...557 (with n 5s).

Original entry on oeis.org

2, 1, 1, 2, 1, 3, 3, 3, 1, 3, 2, 5, 3, 1, 4, 3, 3, 3, 4, 2, 1, 2, 7, 4, 4, 4, 4, 5, 4, 4, 3, 2, 3, 5, 3, 4, 3, 7, 2, 4, 2, 4, 4, 8, 2, 4, 3, 6, 3, 2, 4, 7, 7, 7, 5, 5, 6, 7, 2, 4, 6, 3, 5, 5, 2, 5, 2, 4, 5, 5, 3, 9, 10, 6, 5, 6, 4, 4, 4, 5, 4, 5, 3, 6, 6, 4, 1
Offset: 1

Views

Author

Parthasarathy Nambi, Apr 20 2005

Keywords

Examples

			The number of distinct prime divisors of 57 is 2.
The number of distinct prime divisors of 557 is 1 (prime).
The number of distinct prime divisors of 5557 is 1 (prime).
		

Crossrefs

Programs

  • Maple
    A104524 := proc(n) local x ;x := [7,seq(5,k=1..n)] ; add(op(i,x)*10^(i-1),i=1..nops(x)) ; numtheory[factorset](%) ; nops(%) ; end proc: # R. J. Mathar, Aug 24 2011
  • Mathematica
    Table[PrimeNu[(50*10^n + 13)/9], {n, 1, 50}](* G. C. Greubel, May 07 2017 *)
    Table[PrimeNu[FromDigits[PadLeft[{7},n,5]]],{n,2,90}] (* Harvey P. Dale, Dec 03 2021 *)
  • PARI
    a(n) = omega((5*10^(n+1)+13)/9); \\ Michel Marcus, May 08 2017

Formula

a(n) = A001221(A178769(n+1)). - R. J. Mathar, Aug 24 2011

Extensions

More terms from Michel Marcus, May 08 2017
More terms from Amiram Eldar, Jan 25 2020

A309613 Digits of the 10-adic integer (13/9)^(1/3).

Original entry on oeis.org

3, 9, 6, 9, 6, 6, 2, 5, 6, 6, 5, 7, 4, 8, 2, 6, 1, 5, 2, 3, 5, 4, 9, 5, 1, 2, 1, 5, 3, 6, 9, 7, 1, 6, 7, 5, 5, 3, 7, 6, 4, 9, 5, 3, 1, 7, 9, 8, 0, 4, 4, 5, 7, 1, 2, 1, 3, 3, 4, 9, 1, 0, 1, 7, 6, 1, 9, 8, 0, 4, 1, 3, 7, 6, 7, 1, 2, 2, 0, 1, 1, 5, 4, 9, 2, 5, 2, 8, 9, 7, 5, 0, 9, 1, 4, 5, 4, 9, 7, 3
Offset: 0

Views

Author

Seiichi Manyama, Aug 10 2019

Keywords

Examples

			       3^3 == 7      (mod 10).
      93^3 == 57     (mod 10^2).
     693^3 == 557    (mod 10^3).
    9693^3 == 5557   (mod 10^4).
   69693^3 == 55557  (mod 10^5).
  669693^3 == 555557 (mod 10^6).
		

Crossrefs

Programs

  • PARI
    N=100; Vecrev(digits(lift(chinese(Mod((13/9+O(2^N))^(1/3), 2^N), Mod((13/9+O(5^N))^(1/3), 5^N)))), N)
    
  • Ruby
    def A309613(n)
      ary = [3]
      a = 3
      n.times{|i|
        b = (a + 3 * (9 * a ** 3 - 13)) % (10 ** (i + 2))
        ary << (b - a) / (10 ** (i + 1))
        a = b
      }
      ary
    end
    p A309613(100)

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 3 * (9 * b(n-1)^3 - 13) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.

A173735 a(n) = (10^n + 26)/9.

Original entry on oeis.org

3, 4, 14, 114, 1114, 11114, 111114, 1111114, 11111114, 111111114, 1111111114, 11111111114, 111111111114, 1111111111114, 11111111111114, 111111111111114, 1111111111111114, 11111111111111114, 111111111111111114, 1111111111111111114, 11111111111111111114, 111111111111111111114
Offset: 0

Views

Author

Vincenzo Librandi, Feb 23 2010

Keywords

Crossrefs

Programs

Formula

a(n) = a(n-1) + 10^(n-1) = 10*a(n-1) - 26, a(0)=3.
a(n) = 11*a(n-1) - 10*a(n-2). - Vincenzo Librandi, Jul 05 2012
a(n) = 2*(A002279(n-1) + 2). - Martin Ettl, Nov 12 2012
G.f.: (3-29*x)/((1-x)*(1-10*x)). - Ivan Panchenko, Nov 05 2013
From Elmo R. Oliveira, Jun 18 2025: (Start)
E.g.f.: exp(x)*(26 + exp(9*x))/9.
a(n) = 2*A178769(n-1) for n >= 1. (End)

Extensions

a(0) from Ivan Panchenko, Nov 05 2013
Showing 1-3 of 3 results.