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A321072
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One of the two successive approximations up to 11^n for 11-adic integer sqrt(3). Here the 5 (mod 11) case (except for n = 0).
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3
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0, 5, 27, 753, 11401, 26042, 1475501, 17419550, 95368234, 738444877, 21959974096, 73834823298, 2356328188186, 11771613318349, 149862461894073, 3567610964143242, 7744859133558893, 421292427905708342, 1937633513403589655, 18617385453880284098, 18617385453880284098
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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 solution to x^2 == 3 (mod 11^n) in the range [0, 11^n - 1] and congruent to 5 modulo 11.
Differs from A034946 since a(20). A034946 lists terms of this sequence without repetition.
A321073 is the approximation (congruent to 6 mod 11) of another square root of 3 over the 11-adic field.
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LINKS
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FORMULA
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For n > 0, a(n) = 11^n - A321073(n).
a(n) = Sum_{i=0..n-1} A321074(i)*11^i.
a(n) == ((5 + sqrt(21))/2)^(11^n) + ((5 - sqrt(21))/2)^(11^n) (mod 11^n). - Peter Bala, Dec 04 2022
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EXAMPLE
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5^2 = 25 = 3 + 2*11.
27^2 = 729 = 3 + 6*11^2.
753^2 = 567009 = 3 + 426*11^3.
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PROG
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(PARI) a(n) = truncate(sqrt(3+O(11^n)))
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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