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.

A093168 Primes of the form 30*R_k + 7, where R_k is the repunit (A002275) of length k.

Original entry on oeis.org

7, 37, 337, 333337, 3333333333333333333333333333333333333333333337
Offset: 1

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Author

Rick L. Shepherd, Mar 26 2004

Keywords

Comments

Primes of the form (3*10^k + 33)/9. - Vincenzo Librandi, Nov 16 2010

Crossrefs

Cf. A056680 (corresponding k), A099411, A173766.

A104483 Number of distinct prime divisors of 33...337 (with n 3s).

Original entry on oeis.org

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

Views

Author

Parthasarathy Nambi, Apr 18 2005

Keywords

Comments

Number of distinct prime factors of (10^(n + 1) - 1)/3 + 4. - Stefan Steinerberger, Mar 06 2006

Examples

			The number of distinct prime divisors of 37 is 1 (prime).
The number of distinct prime divisors of 337 is 1 (prime).
The number of distinct prime divisors of 3337 is 2.
		

Crossrefs

Programs

  • Mathematica
    Table[Length[FactorInteger[(10^(n + 1) - 1)/3 + 4]], {n, 1, 50}] (* Stefan Steinerberger, Mar 06 2006 *)

Formula

a(n) = A001221(A173766(n+1)). - Amiram Eldar, Jan 24 2020

Extensions

More terms from Stefan Steinerberger, Mar 06 2006
More terms from Amiram Eldar, Jan 24 2020

A178769 a(n) = (5*10^n + 13)/9.

Original entry on oeis.org

2, 7, 57, 557, 5557, 55557, 555557, 5555557, 55555557, 555555557, 5555555557, 55555555557, 555555555557, 5555555555557, 55555555555557, 555555555555557, 5555555555555557, 55555555555555557, 555555555555555557, 5555555555555555557, 55555555555555555557, 555555555555555555557
Offset: 0

Views

Author

Bruno Berselli, Jun 13 2010

Keywords

Crossrefs

Cf. A165246 (..17, 117, 1117,..), A173193 (..27, 227, 2227,..), A173766 (..37, 337, 3337,..), A173772 (..47, 447, 4447,..), A067275 (..67, 667, 6667,..), A002281 (..77, 777, 7777,..), A173812 (..87, 887, 8887,..), A173833 (..97, 997, 9997,..).
Cf. A093143.

Programs

  • GAP
    List([0..20], n -> (5*10^n+13)/9); # G. C. Greubel, Jan 24 2019
  • Magma
    [(5*10^n+13)/9: n in [0..20]]; // Vincenzo Librandi, Jun 06 2013
    
  • Mathematica
    CoefficientList[Series[(2 - 15 x) / ((1 - x) (1 - 10 x)), {x, 0, 20}], x] (* Vincenzo Librandi, Jun 06 2013 *)
    LinearRecurrence[{11,-10},{2,7},20] (* Harvey P. Dale, Feb 28 2017 *)
  • PARI
    vector(20, n, n--; (5*10^n+13)/9) \\ G. C. Greubel, Jan 24 2019
    
  • Sage
    [(5*10^n+13)/9 for n in (0..20)] # G. C. Greubel, Jan 24 2019
    

Formula

a(n)^(4*k+2) + 1 == 0 (mod 250) for n > 1, k >= 0.
G.f.: (2-15*x)/((1-x)*(1-10*x)).
a(n) - 11*a(n-1) + 10*a(n-2) = 0 (n > 1).
a(n) = a(n-1) + 5*10^(n-1) = 10*a(n-1) - 13 for n > 0.
a(n) = 1 + Sum_{i=0..n} A093143(i). - Bruno Berselli, Feb 16 2015
E.g.f.: exp(x)*(5*exp(9*x) + 13)/9. - Elmo R. Oliveira, Sep 09 2024

A309569 Digits of the 10-adic integer (11/3)^(1/3).

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Aug 10 2019

Keywords

Examples

			       3^3 == 7      (mod 10).
      33^3 == 37     (mod 10^2).
     233^3 == 337    (mod 10^3).
    2233^3 == 3337   (mod 10^4).
   52233^3 == 33337  (mod 10^5).
  852233^3 == 333337 (mod 10^6).
		

Crossrefs

Programs

  • PARI
    N=100; Vecrev(digits(lift(chinese(Mod((11/3+O(2^N))^(1/3), 2^N), Mod((11/3+O(5^N))^(1/3), 5^N)))), N)
    
  • Ruby
    def A309569(n)
      ary = [3]
      a = 3
      n.times{|i|
        b = (a + 9 * (3 * a ** 3 - 11)) % (10 ** (i + 2))
        ary << (b - a) / (10 ** (i + 1))
        a = b
      }
      ary
    end
    p A309569(100)

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 9 * (3 * b(n-1)^3 - 11) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.
Showing 1-4 of 4 results.