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A322931
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Digits of the 8-adic integer 3^(1/3).
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4
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3, 7, 5, 0, 7, 3, 0, 1, 7, 1, 7, 6, 4, 2, 3, 6, 7, 7, 7, 0, 3, 1, 2, 0, 1, 7, 2, 6, 1, 2, 5, 4, 1, 1, 1, 2, 3, 5, 5, 2, 3, 5, 4, 7, 3, 6, 0, 0, 3, 4, 7, 1, 3, 3, 6, 4, 6, 0, 0, 4, 4, 6, 0, 5, 6, 4, 1, 5, 5, 6, 0, 0, 0, 6, 2, 6, 1, 0, 7, 1, 6, 0, 0, 4, 6, 5, 0, 7, 0, 1, 3, 7, 3, 7, 0, 0, 2, 0, 6, 1
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OFFSET
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0,1
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COMMENTS
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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) + 5 * (b(n-1)^3 - 3) mod 8^n for n > 1, then a(n) = (b(n+1) - b(n))/8^n. - Seiichi Manyama, Aug 14 2019
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EXAMPLE
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4671710370573^3 == 3 (mod 8^13) in octal.
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PROG
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(PARI) N=100; Vecrev(digits(lift((3+O(2^(3*N)))^(1/3)), 8), N) \\ Seiichi Manyama, Aug 14 2019
(Ruby)
ary = [3]
a = 3
n.times{|i|
b = (a + 5 * (a ** 3 - 3)) % (8 ** (i + 2))
ary << (b - a) / (8 ** (i + 1))
a = b
}
ary
end
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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