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A225411
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10-adic integer x such that x^3 == (6*(10^(n+1)-1)/9)+1 mod 10^(n+1) for all n.
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10
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3, 2, 5, 3, 5, 5, 4, 1, 6, 5, 5, 3, 1, 6, 8, 4, 8, 9, 4, 8, 3, 0, 3, 8, 6, 8, 6, 2, 7, 7, 5, 3, 5, 5, 6, 1, 6, 4, 6, 6, 4, 3, 7, 4, 7, 6, 7, 1, 3, 6, 5, 8, 1, 0, 3, 2, 9, 5, 1, 6, 3, 2, 8, 8, 3, 3, 4, 7, 9, 2, 5, 9, 0, 7, 1, 9, 6, 9, 6, 9, 2, 6, 0, 2, 0, 2, 6, 5, 3, 4, 6, 8, 6, 9, 9, 1, 2, 5, 4, 2
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OFFSET
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0,1
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COMMENTS
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This is the 10's complement of A225402.
Equivalently, the 10-adic cube root of 1/3, i.e., solution to 3*x^3 = 1 (mod 10^n) for all n. - M. F. Hasler, Jan 02 2019
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LINKS
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FORMULA
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Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 9 * (3 * b(n-1)^3 - 1) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 12 2019
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EXAMPLE
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3^3 == 7 (mod 10).
23^3 == 67 (mod 10^2).
523^3 == 667 (mod 10^3).
3523^3 == 6667 (mod 10^4).
53523^3 == 66667 (mod 10^5).
553523^3 == 666667 (mod 10^6).
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MAPLE
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op([1, 3], padic:-rootp(3*x^3 -1, 10, 101)); # Robert Israel, Aug 04 2019
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PROG
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(PARI) n=0; for(i=1, 100, m=(2*(10^i-1)/3)+1; for(x=0, 9, if(((n+(x*10^(i-1)))^3)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break)))
(PARI) upto(N=100, m=1/3)=Vecrev(digits(lift(chinese(Mod((m+O(5^N))^m, 5^N), Mod((m+O(2^N))^m, 2^N)))), N) \\ Following Andrew Howroyd's code for A319740. - M. F. Hasler, Jan 02 2019
(Ruby)
ary = [3]
a = 3
n.times{|i|
b = (a + 9 * (3 * a ** 3 - 1)) % (10 ** (i + 2))
ary << (b - a) / (10 ** (i + 1))
a = b
}
ary
end
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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