A321072 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).

0, 5, 27, 753, 11401, 26042, 1475501, 17419550, 95368234, 738444877, 21959974096, 73834823298, 2356328188186, 11771613318349, 149862461894073, 3567610964143242, 7744859133558893, 421292427905708342, 1937633513403589655, 18617385453880284098, 18617385453880284098

Offset: 0

Author

Jianing Song, Oct 27 2018

Keywords

• nonn
• easy

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 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.

Examples

			5^2 = 25 = 3 + 2*11.
27^2 = 729 = 3 + 6*11^2.
753^2 = 567009 = 3 + 426*11^3.
		

Links

• Wikipedia, p-adic number

Crossrefs

Cf. A034946, A321073, A321074.

Programs

PARI

a(n) = truncate(sqrt(3+O(11^n)))

Formula

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