A240162 Tower of 3's modulo n.
0, 1, 0, 3, 2, 3, 6, 3, 0, 7, 9, 3, 1, 13, 12, 11, 7, 9, 18, 7, 6, 9, 18, 3, 12, 1, 0, 27, 10, 27, 23, 27, 9, 7, 27, 27, 36, 37, 27, 27, 27, 27, 2, 31, 27, 41, 6, 27, 6, 37, 24, 27, 50, 27, 42, 27, 18, 39, 49, 27, 52, 23, 27, 59, 27, 9, 52, 7, 18, 27, 49, 27
Offset: 1
Examples
a(7) = 6. For any natural number X, 3^X is a positive odd multiple of 3. 3^(any positive odd multiple of three) mod 7 is always 6. a(9) = 0, since 3^(3^X) is divisible by 9 for any natural number X. In our case, X itself is a tower of 3's. a(100000000) = 64195387, giving the rightmost eight digits of Graham's Number. From _Robert Munafo_, Apr 19 2020: (Start) a(1) = 0, because 3 mod 1 = 0. a(2) = 1, because 3^3 mod 2 = 1. a(3) = 0, because 3^3^3 mod 3 = 0. a(4) = 3, because 3^3^3^3 = 3^N for odd N, 3^N = 3 mod 4 for all odd N. a(5) = 3^3^3^3^3 mod 5, and we should look at the sequence 3^N mod 5. We find that 3^N = 2 mod 5 whenever N = 3 mod 4. As just shown in the a(4) example, 3^3^3^3 = 3 mod 4. (End)
Links
- Wayne VanWeerthuizen, Table of n, a(n) for n = 1..10000
Programs
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Haskell
import Math.NumberTheory.Moduli (powerMod) a245972 n = powerMod 3 (a245972 $ a000010 n) n -- Reinhard Zumkeller, Feb 01 2015
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Maple
A:= proc(n) option remember; 3 &^ A(numtheory:-phi(n)) mod n end proc: A(2):= 1; seq(A(n), n=2..100); # Robert Israel, Aug 01 2014
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Mathematica
a[1] = 0; a[n_] := a[n] = PowerMod[3, a[EulerPhi[n]], n]; Array[a, 72] (* Jean-François Alcover, Feb 09 2018 *)
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Sage
def A(n): if ( n <= 10 ): return 27%n else: return power_mod(3,A(euler_phi(n)),n)
Formula
a(n) = 3^a(A000010(n)) mod n. - Robert Israel, Aug 01 2014
Comments