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A319098
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).
11
0, 5, 40, 138, 824, 3225, 87260, 793154, 793154, 29617159, 191031587, 1320932583, 7252912812, 7252912812, 7252912812, 2041922131359, 16284606661188, 82750467800390, 1013272523749218, 9155340513301463, 31953130884047749, 111745397181659750, 670291261264943757
OFFSET
0,2
COMMENTS
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.
FORMULA
a(n) = A319097(n)*(A212153(n)-1) mod 7^n = A319097(n)*A212153(n)^2 mod 7^n.
a(n) = A319199(n)*(A210852(n)-1) mod 7^n = A319199(n)*A210852(n)^2 mod 7^n.
EXAMPLE
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.
PROG
(PARI) a(n) = lift(sqrtn(6+O(7^n), 3) * (-1-sqrt(-3+O(7^n)))/2)
CROSSREFS
Approximations of p-adic cubic roots:
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A309444 (5-adic, 4^(1/3));
A319097, this sequence, A319199 (7-adic, 6^(1/3));
A320914, A320915, A321105 (13-adic, 5^(1/3)).
Sequence in context: A153795 A015874 A244725 * A209346 A027264 A025214
KEYWORD
nonn
AUTHOR
Jianing Song, Aug 27 2019
STATUS
approved