A321072 -id:A321072 - cp's OEIS frontend

A321074 Digits of one of the two 11-adic integers sqrt(3).

5, 2, 6, 8, 1, 9, 9, 4, 3, 9, 2, 8, 3, 4, 9, 1, 9, 3, 3, 0, 5, 5, 0, 9, 8, 4, 1, 9, 6, 9, 3, 0, 7, 5, 8, 6, 3, 9, 0, 9, 7, 7, 9, 8, 10, 5, 8, 6, 9, 3, 5, 9, 4, 7, 2, 1, 1, 0, 1, 0, 8, 1, 6, 5, 7, 10, 8, 2, 4, 7, 8, 7, 2, 3, 3, 1, 10, 6, 0, 10, 0, 6, 2, 5, 1, 10, 3

Offset: 0

Jianing Song, Oct 27 2018

This square root of 3 in the 11-adic field ends with digit 5. The other, A321075, ends with digit 6.

			...9093685703969148905503391943829349918625.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Peter Bala, Using Chebyshev polynomials to find the p-adic square roots of 2 and 3, Dec 2022.
Wikipedia, p-adic number

Cf. A321072, A321075, A322087, A322088.

a(n) = truncate(sqrt(3+O(11^(n+1))))\11^n

seq(n)={Vecrev(digits(truncate(sqrt(3 + O(11^n))), 11), n)} \\ Andrew Howroyd, Nov 03 2018

a(n) = (A321072(n+1) - A321072(n))/11^n.
For n > 0, a(n) = 10 - A321075(n).
This 11-adic integer equals the 11-adic limit as n -> oo of 2*T(11^n,5/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Dec 05 2022

A034946 Successive approximations to 11-adic integer sqrt(3).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

nonn

Terms of A321072 without repetition. - Jianing Song, Oct 31 2018

K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973, p. 35.

Jianing Song, Table of n, a(n) for n = 0..543

Cf. A321072.

PARI
```
sqrt(3+O(11^40))
```

A321072(n) = truncate(sqrt(3+O(11^n)))
for(n=0, 50, if(valuation(A321072(n)^2-3, 11)!=valuation(A321072(n+1)^2-3, 11), print1(A321072(n), ", "))) \\ Jianing Song, Nov 01 2018

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).

Original entry on oeis.org

0, 6, 94, 578, 3240, 135009, 296060, 2067621, 118990647, 1619502814, 3977450505, 211476847313, 782100188535, 22751098825582, 229887371689168, 609637205272409, 38204870730013268, 84154600593585429, 3622283800088641826, 42541704994534262193, 654132609478679725103

Offset: 0

Jianing Song, Oct 27 2018

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.

			6^2 = 36 = 3 + 3*11.
94^2 = 8836 = 3 + 73*11^2.
578^2 = 334084 = 3 + 251*11^3.

Wikipedia, p-adic number

Cf. A321072, A321075.

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

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

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A321074 Digits of one of the two 11-adic integers sqrt(3).

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A034946 Successive approximations to 11-adic integer sqrt(3).

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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).

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