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A319098
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One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 5 (mod 7) case (except for n = 0).
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11
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0, 5, 40, 138, 824, 3225, 87260, 793154, 793154, 29617159, 191031587, 1320932583, 7252912812, 7252912812, 7252912812, 2041922131359, 16284606661188, 82750467800390, 1013272523749218, 9155340513301463, 31953130884047749, 111745397181659750, 670291261264943757
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OFFSET
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0,2
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COMMENTS
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For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 5 mod 7 such that k^3 - 6 is divisible by 7^n.
For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 7). If k is a cube in 7-adic field, then k has exactly three cubic roots.
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LINKS
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FORMULA
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EXAMPLE
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The unique number k in [1, 7^2] and congruent to 5 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 40, so a(2) = 40.
The unique number k in [1, 7^3] and congruent to 5 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 138, so a(3) = 138.
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PROG
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(PARI) a(n) = lift(sqrtn(6+O(7^n), 3) * (-1-sqrt(-3+O(7^n)))/2)
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CROSSREFS
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Approximations of p-adic cubic roots:
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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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