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 1-5 of 5 results.

A309699 Digits of the 6-adic integer 5^(1/5).

Original entry on oeis.org

5, 4, 0, 3, 1, 5, 0, 0, 3, 3, 2, 1, 3, 0, 0, 3, 4, 3, 1, 1, 1, 1, 1, 4, 3, 4, 0, 5, 3, 1, 1, 5, 3, 3, 0, 2, 2, 2, 5, 3, 5, 5, 2, 5, 2, 2, 2, 3, 4, 2, 0, 5, 4, 3, 3, 2, 0, 0, 4, 1, 1, 5, 5, 5, 0, 0, 1, 4, 3, 5, 4, 5, 1, 5, 5, 0, 5, 4, 0, 4, 4, 4, 4, 3, 4, 4, 0, 4, 3, 4, 0, 5, 4, 4
Offset: 0

Views

Author

Seiichi Manyama, Aug 13 2019

Keywords

Comments

x = ...513045,
x^2 = ...433521,
x^3 = ...051525,
x^4 = ...354241,
x^5 = ...000005.

Crossrefs

Digits of the k-adic integer (k-1)^(1/(k-1)): A309698 (k=4), this sequence (k=6), A309700 (k=8), A225458 (k=10).
Cf. A309448.

Programs

  • PARI
    N=100; Vecrev(digits(lift(chinese(Mod((5+O(2^N))^(1/5), 2^N), Mod((5+O(3^N))^(1/5), 3^N))), 6), N)
    
  • Ruby
    def A309699(n)
      ary = [5]
      a = 5
      n.times{|i|
        b = (a + a ** 5 - 5) % (6 ** (i + 2))
        ary << (b - a) / (6 ** (i + 1))
        a = b
      }
      ary
    end
    p A309699(100)

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 5, b(n) = b(n-1) + b(n-1)^5 - 5 mod 6^n for n > 1, then a(n) = (b(n+1) - b(n))/6^n.

A225458 10-adic integer x such that x^9 = 9.

Original entry on oeis.org

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

Views

Author

Aswini Vaidyanathan, May 11 2013

Keywords

Examples

			       9^9 == 9 (mod 10).
      89^9 == 9 (mod 10^2).
     289^9 == 9 (mod 10^3).
    1289^9 == 9 (mod 10^4).
   21289^9 == 9 (mod 10^5).
  921289^9 == 9 (mod 10^6).
		

Crossrefs

Digits of the k-adic integer (k-1)^(1/(k-1)): A309698 (k=4), A309699 (k=6), A309700 (k=8), this sequence (k=10).

Programs

  • PARI
    n=0;for(i=1,100,m=9;for(x=0,9,if(((n+(x*10^(i-1)))^9)%(10^i)==m,n=n+(x*10^(i-1));print1(x", ");break)))
    
  • PARI
    N=100; Vecrev(digits(lift(chinese(Mod((9+O(2^N))^(1/9), 2^N), Mod((9+O(5^N))^(1/9), 5^N)))), N) \\ Seiichi Manyama, Aug 06 2019
    
  • Ruby
    def A225458(n)
      ary = [9]
      a = 9
      n.times{|i|
        b = (a + a ** 9 - 9) % (10 ** (i + 2))
        ary << (b - a) / (10 ** (i + 1))
        a = b
      }
      ary
    end
    p A225458(100) # Seiichi Manyama, Aug 14 2019

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 9, b(n) = b(n-1) + b(n-1)^9 - 9 mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 14 2019

A309700 Digits of the 8-adic integer 7^(1/7).

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Aug 13 2019

Keywords

Crossrefs

Digits of the k-adic integer (k-1)^(1/(k-1)): A309698 (k=4), A309699 (k=6), this sequence (k=8), A225458 (k=10).
Cf. A225445.

Programs

  • PARI
    N=100; Vecrev(digits(lift((7+O(2^(3*N)))^(1/7)), 8), N)
    
  • Ruby
    def A309700(n)
      ary = [7]
      a = 7
      n.times{|i|
        b = (a + a ** 7 - 7) % (8 ** (i + 2))
        ary << (b - a) / (8 ** (i + 1))
        a = b
      }
      ary
    end
    p A309700(100)

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 7, b(n) = b(n-1) + b(n-1)^7 - 7 mod 8^n for n > 1, then a(n) = (b(n+1) - b(n))/8^n.

A322931 Digits of the 8-adic integer 3^(1/3).

Original entry on oeis.org

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

Views

Author

Patrick A. Thomas, Dec 31 2018

Keywords

Comments

The octal version of A225404.

Examples

			4671710370573^3 == 3 (mod 8^13) in octal.
		

Crossrefs

Cf. A225404 (decimal version), A290563, A309698, A322932, A322933.

Programs

  • PARI
    N=100; Vecrev(digits(lift((3+O(2^(3*N)))^(1/3)), 8), N) \\ Seiichi Manyama, Aug 14 2019
    
  • Ruby
    def A322931(n)
      ary = [3]
      a = 3
      n.times{|i|
        b = (a + 5 * (a ** 3 - 3)) % (8 ** (i + 2))
        ary << (b - a) / (8 ** (i + 1))
        a = b
      }
      ary
    end
    p A322931(100) # Seiichi Manyama, Aug 14 2019

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 5 * (b(n-1)^3 - 3) mod 8^n for n > 1, then a(n) = (b(n+1) - b(n))/8^n. - Seiichi Manyama, Aug 14 2019

A309722 Digits of the 4-adic integer (1/3)^(1/3).

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Aug 14 2019

Keywords

Crossrefs

Digits of the k-adic integer (1/(k-1))^(1/(k-1)): this sequence (k=4), A309723 (k=6), A309724 (k=8), A225464 (k=10).

Programs

  • PARI
    N=100; Vecrev(digits(lift((1/3+O(2^(2*N)))^(1/3)), 4), N)
    
  • Ruby
    def A309722(n)
      ary = [3]
      a = 3
      n.times{|i|
        b = (a + 3 * (3 * a ** 3 - 1)) % (4 ** (i + 2))
        ary << (b - a) / (4 ** (i + 1))
        a = b
      }
      ary
    end
    p A309722(100)

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 3 * (3 * b(n-1)^3 - 1) mod 4^n for n > 1, then a(n) = (b(n+1) - b(n))/4^n.
Showing 1-5 of 5 results.