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.

A309698 Digits of the 4-adic integer 3^(1/3).

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Aug 13 2019

Keywords

Crossrefs

Digits of the k-adic integer (k-1)^(1/(k-1)): this sequence (k=4), A309699 (k=6), A309700 (k=8), A225458 (k=10).

Programs

  • PARI
    N=100; Vecrev(digits(lift((3+O(2^(2*N)))^(1/3)), 4), N)
    
  • Ruby
    def A309698(n)
      ary = [3]
      a = 3
      n.times{|i|
        b = (a + a ** 3 - 3) % (4 ** (i + 2))
        ary << (b - a) / (4 ** (i + 1))
        a = b
      }
      ary
    end
    p A309698(100)

Formula

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

A322932 Digits of the 8-adic integer 5^(1/3).

Original entry on oeis.org

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

Views

Author

Patrick A. Thomas, Dec 31 2018

Keywords

Examples

			20447135^3 == 5 (mod 8^8) in octal.
		

Crossrefs

Programs

  • PARI
    N=100; Vecrev(digits(lift((5+O(2^(3*N)))^(1/3)), 8), N) \\ Seiichi Manyama, Aug 14 2019
    
  • Ruby
    def A322932(n)
      ary = [5]
      a = 5
      n.times{|i|
        b = (a + 5 * (a ** 3 - 5)) % (8 ** (i + 2))
        ary << (b - a) / (8 ** (i + 1))
        a = b
      }
      ary
    end
    p A322932(100) # Seiichi Manyama, Aug 14 2019

Formula

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

A322933 Digits of the 8-adic integer 7^(1/3).

Original entry on oeis.org

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

Views

Author

Patrick A. Thomas, Dec 31 2018

Keywords

Comments

The octal version of A225405.

Examples

			56027726627^3 == 7 (mod 8^11) in octal.
		

Crossrefs

Cf. A225405 (decimal version), A322931, A322932.

Programs

  • PARI
    N=100; Vecrev(digits(lift((7+O(2^(3*N)))^(1/3)), 8), N) \\ Seiichi Manyama, Aug 14 2019
    
  • Ruby
    def A322933(n)
      ary = [7]
      a = 7
      n.times{|i|
        b = (a + 5 * (a ** 3 - 7)) % (8 ** (i + 2))
        ary << (b - a) / (8 ** (i + 1))
        a = b
      }
      ary
    end
    p A322933(100) # Seiichi Manyama, Aug 14 2019

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 7, b(n) = b(n-1) + 5 * (b(n-1)^3 - 7) mod 8^n for n > 1, then a(n) = (b(n+1) - b(n))/8^n. - Seiichi Manyama, Aug 14 2019
Showing 1-3 of 3 results.