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

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

Author

Jianing Song, Oct 27 2018

Keywords

• nonn

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.
		

Links

• Jianing Song, Table of n, a(n) for n = 2..1000
• Wikipedia, p-adic number

Crossrefs

Cf. A321080, A321083, A321690, A321691.

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.