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.

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A321080 Approximations up to 2^n for 2-adic integer log_5(-3).

Original entry on oeis.org

0, 1, 3, 3, 3, 3, 35, 35, 163, 163, 675, 1699, 1699, 1699, 9891, 9891, 42659, 42659, 42659, 304803, 304803, 304803, 304803, 4499107, 4499107, 21276323, 21276323, 21276323, 155494051, 423929507, 423929507, 1497671331, 1497671331, 1497671331, 10087605923
Offset: 2

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Author

Jianing Song, Oct 27 2018

Keywords

Comments

a(n) is the unique number x in [0, 2^(n-2) - 1] such that 5^x == -3 (mod 2^n). This is well defined because {5^x mod 2^n : 0 <= x <= 2^(n-2) - 1} = {1, 5, 9, ..., 2^n - 3}.
For any odd 2-adic integer x, define log(x) = -Sum_{k>=1} (1 - x)^k/k (which always converges over the 2-adic field) and log_x(y) = log(y)/log(x), then we have log(-1) = 0. If we further define exp(x) = Sum_{k>=0} x^k/k! for 2-adic integers divisible by 4, then we have exp(log(x)) = x if and only if x == 1 (mod 4). As a result, log_5(-3) = log_5(3) = log_(-5)(3) = log_(-5)(-3), but it's better to be stated as log_5(-3).
For n > 0, a(n) is also the unique number x in [0, 2^(n-2) - 1] such that (-5)^x == 3 (mod 2^n).
a(n) is the multiplicative inverse of A321082(n) modulo 2^(n-2).

Examples

			The only number in the range [0, 2^(n-2) - 1] for n = 2 is 0, so a(2) = 0.
5^a(2) + 3 = 4 which is not divisible by 8, so a(3) = a(2) + 2^0 = 1.
5^a(3) + 3 = 8 which is not divisible by 16, so a(4) = a(3) + 2^1 = 3.
5^a(4) + 3 = 128 which is divisible by 32, 64 and 128 but not 256, so a(5) = a(6) = a(7) = a(4) = 3, a(8) = a(7) + 2^5 = 35.
5^a(8) + 3 = ... which is divisible by 512 but not 1024, so a(9) = a(8) = 35, a(10) = a(9) + 2^7 = 163.
		

Crossrefs

Programs

  • PARI
    b(n) = {my(v=vector(n)); for(n=3, n, v[n] = v[n-1] + if(Mod(5,2^n)^v[n-1] + 3==0, 0, 2^(n-3))); v}
    a(n) = b(n)[n] \\ Program provided by Andrew Howroyd
    
  • PARI
    a(n)={if(n<3, 0, truncate(log(-3 + O(2^n))/log(5 + O(2^n))))} \\ Andrew Howroyd, Nov 03 2018

Formula

a(2) = 0; for n >= 3, a(n) = a(n-1) if 5^a(n-1) + 3 is divisible by 2^n, otherwise a(n-1) + 2^(n-3).
a(n) = Sum_{i=0..n-3} A321081(i)*2^i (empty sum yields 0 for n = 2).
a(n) = A321691(n+2)/A321690(n+2) mod 2^n.

A321083 Digits of the 2-adic integer log_(-3)(5).

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0
Offset: 0

Views

Author

Jianing Song, Oct 27 2018

Keywords

Comments

See A321082 for the definition of log_5(-3) and more information.
Multiplicative inverse of A321081.

Examples

			log_(-3)(5) = ...0110111111010101110010000011110100001011.
		

Crossrefs

Programs

  • PARI
    b(n) = {my(v=vector(n)); v[3]=1; for(n=4, n, v[n] = v[n-1] + if(Mod(-3,2^n)^v[n-1] - 5==0, 0, 2^(n-3))); v}
    a(n) = b(n+3)[n+3]\2^n

Formula

a(n) = 0 if (-3)^A321082(n+2) - 5 is divisible by 2^(n+3), otherwise 1.
Equals to A152228/A321694.
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