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A321080
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Approximations up to 2^n for 2-adic integer log_5(-3).
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3
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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
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OFFSET
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2,3
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COMMENTS
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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).
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LINKS
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FORMULA
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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).
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EXAMPLE
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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.
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PROG
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(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}
(PARI) a(n)={if(n<3, 0, truncate(log(-3 + O(2^n))/log(5 + O(2^n))))} \\ Andrew Howroyd, Nov 03 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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