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A324083 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 10 (mod 13) case (except for n = 0). 13
0, 10, 101, 1622, 14804, 214731, 214731, 9868349, 637353519, 637353519, 637353519, 552071320915, 552071320915, 23850156443396, 1538225689404661, 48786742317796129, 560645672458703699, 5218561936740962586, 13868977856122300519, 126324384808079693648 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 10 mod 13 such that k^4 - 3 is divisible by 13^n.
For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.
LINKS
Wikipedia, p-adic number
FORMULA
a(n) = A324077(n)*A286840(n) mod 13^n = A324084(n)*A286841(n) mod 13^n.
For n > 0, a(n) = 13^n - A324082(n).
a(n)^2 == A322086(n) (mod 13^n).
EXAMPLE
The unique number k in [1, 13^2] and congruent to 10 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 101, so a(2) = 101.
The unique number k in [1, 13^3] and congruent to 10 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 1622, so a(3) = 1622.
PROG
(PARI) a(n) = lift(-sqrtn(3+O(13^n), 4))
CROSSREFS
Sequence in context: A287014 A108892 A309802 * A162849 A041182 A036299
KEYWORD
nonn
AUTHOR
Jianing Song, Sep 01 2019
STATUS
approved

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Last modified April 18 21:51 EDT 2024. Contains 371781 sequences. (Running on oeis4.)