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 1-2 of 2 results.

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.

A225448 10-adic integer x such that x^7 = -7.

Original entry on oeis.org

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

Views

Author

Aswini Vaidyanathan, May 11 2013

Keywords

Comments

This is the 10's complement of A225445.

Examples

			       7^7 == -7 (mod 10).
      57^7 == -7 (mod 10^2).
     657^7 == -7 (mod 10^3).
    7657^7 == -7 (mod 10^4).
   87657^7 == -7 (mod 10^5).
  987657^7 == -7 (mod 10^6).
		

Crossrefs

Cf. A225445.

Programs

  • PARI
    n=0;for(i=1,100,m=(10^i-7);for(x=0,9,if(((n+(x*10^(i-1)))^7)%(10^i)==m,n=n+(x*10^(i-1));print1(x", ");break)))
    
  • PARI
    N=100; Vecrev(digits(lift(chinese(Mod((-7+O(2^N))^(1/7), 2^N), Mod((-7+O(5^N))^(1/7), 5^N)))), N) \\ Seiichi Manyama, Aug 06 2019
Showing 1-2 of 2 results.