OFFSET
2,3
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).
LINKS
Jianing Song, Table of n, a(n) for n = 2..1000
Wikipedia, p-adic number
FORMULA
EXAMPLE
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.
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Oct 27 2018
STATUS
approved