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A240162
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Tower of 3's modulo n.
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10
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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
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OFFSET
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1,4
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COMMENTS
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a(n) = (3^(3^(3^(3^(3^ ... ))))) mod n, provided sufficient 3's are in the tower such that adding more doesn't affect the value of a(n).
For values of n significantly less than Graham's Number, a(n) is equal to Graham's Number mod n.
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LINKS
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FORMULA
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EXAMPLE
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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.
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)
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MAPLE
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A:= proc(n) option remember; 3 &^ A(numtheory:-phi(n)) mod n end proc:
A(2):= 1;
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MATHEMATICA
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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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PROG
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(Sage)
def A(n):
if ( n <= 10 ):
return 27%n
else:
return power_mod(3, A(euler_phi(n)), n)
(Haskell)
import Math.NumberTheory.Moduli (powerMod)
a245972 n = powerMod 3 (a245972 $ a000010 n) n
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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