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 A321073 One of the two successive approximations up to 11^n for 11-adic integer sqrt(3). Here the 6 (mod 11) case (except for n = 0). 2
 0, 6, 94, 578, 3240, 135009, 296060, 2067621, 118990647, 1619502814, 3977450505, 211476847313, 782100188535, 22751098825582, 229887371689168, 609637205272409, 38204870730013268, 84154600593585429, 3622283800088641826, 42541704994534262193, 654132609478679725103 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS For n > 0, a(n) is the unique solution to x^2 == 3 (mod 11^n) in the range [0, 11^n - 1] and congruent to 6 modulo 11. A321072 is the approximation (congruent to 5 mod 11) of another square root of 3 over the 11-adic field. LINKS Table of n, a(n) for n=0..20. Wikipedia, p-adic number FORMULA For n > 0, a(n) = 11^n - A321072(n). a(n) = Sum_{i=0..n-1} A321075(i)*11^i. a(n) == (3 + sqrt(8))^(11^n) + (3 - sqrt(8))^(11^n) (mod 11^n). - Peter Bala, Dec 04 2022 EXAMPLE 6^2 = 36 = 3 + 3*11. 94^2 = 8836 = 3 + 73*11^2. 578^2 = 334084 = 3 + 251*11^3. PROG (PARI) a(n) = truncate(-sqrt(3+O(11^n))) CROSSREFS Cf. A321072, A321075. Sequence in context: A218682 A078103 A221525 * A198257 A296820 A184983 Adjacent sequences: A321070 A321071 A321072 * A321074 A321075 A321076 KEYWORD nonn,easy AUTHOR Jianing Song, Oct 27 2018 STATUS approved

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Last modified June 8 08:19 EDT 2023. Contains 363157 sequences. (Running on oeis4.)