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.

A321081 Digits of the 2-adic integer log_5(-3).

Original entry on oeis.org

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

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Author

Jianing Song, Oct 27 2018

Keywords

Comments

See A321080 for the definition of log_5(-3) and more information.
Multiplicative inverse of A321083.

Examples

			log_5(-3) = ...1000011001011001010001001010011010100011.
		

Crossrefs

Programs

  • PARI
    b(n) = {my(v=vector(n)); v[2]=0; 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+3)[n+3]\2^n

Formula

a(n) = 0 if 5^A321080(n+2) + 3 is divisible by 2^(n+3), otherwise 1.
Equals to A321694/A152228.

A321082 Approximations up to 2^n for 2-adic integer log_(-3)(5).

Original entry on oeis.org

0, 1, 3, 3, 11, 11, 11, 11, 11, 267, 267, 1291, 3339, 7435, 15627, 15627, 15627, 15627, 15627, 15627, 539915, 539915, 539915, 4734219, 13122827, 29900043, 29900043, 97008907, 97008907, 365444363, 365444363, 1439186187, 3586669835, 7881637131, 16471571723
Offset: 2

Views

Author

Jianing Song, Oct 27 2018

Keywords

Comments

a(n) is the unique number x in [0, 2^(n-2) - 1] such that (-3)^x == 5 (mod 2^n). This is well defined because {(-3)^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_(-3)(5) = log_(-3)(-5) = log_3(5) = log_3(-5), but it's better to be stated as log_(-3)(5).
For n > 0, a(n) is also the unique number x in [0, 2^(n-2) - 1] such that 3^x == -5 (mod 2^n).
a(n) is the multiplicative inverse of A321080(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.
(-3)^a(2) - 5 = -4 which is not divisible by 8, so a(3) = a(2) + 2^0 = 1.
(-3)^a(3) - 5 = -8 which is not divisible by 16, so a(4) = a(3) + 2^1 = 3.
(-3)^a(4) - 5 = -32 which is divisible by 32 but not 64, so a(5) = a(4) = 3, a(6) = a(5) + 2^3 = 11.
(-3)^a(6) - 5 = -177152 which is divisible by 128, 256, 512, 1024 but not 2048, so a(7) = a(8) = a(9) = a(10) = a(6) = 11, a(11) = a(10) + 2^8 = 267.
		

Crossrefs

Programs

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

Formula

a(2) = 0; for n >= 3, a(n) = a(n-1) if (-3)^a(n-1) - 5 is divisible by 2^n, otherwise a(n-1) + 2^(n-3).
a(n) = Sum_{i=0..n-3} A321083(i)*2^i (empty sum yields 0 for n = 2).
a(n) = A321690(n+2)/A321691(n+2) mod 2^n.
Showing 1-2 of 2 results.