A322087 -id:A322087 - cp's OEIS frontend

A324153 Digits of one of the four 13-adic integers 3^(1/4) that is congruent to 11 mod 13.

11, 10, 5, 11, 0, 6, 0, 8, 4, 6, 11, 2, 8, 6, 5, 4, 2, 11, 0, 3, 3, 5, 12, 0, 9, 6, 8, 7, 1, 0, 9, 1, 3, 7, 4, 8, 8, 10, 5, 8, 1, 4, 8, 2, 11, 12, 10, 11, 8, 9, 1, 5, 9, 6, 9, 10, 6, 5, 9, 6, 11, 12, 9, 12, 1, 4, 1, 6, 1, 12, 9, 7, 8, 5, 3, 2, 0, 6, 1, 7, 11

Offset: 0

Jianing Song, Sep 01 2019

One of the two square roots of A322087, where an A-number represents a 13-adic number. The other square root is A324085.

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.

			The unique number k in [1, 13^3] and congruent to 11 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 986 = (5AB)_13, so the first three terms are 11, 10 and 5.

Robert Israel, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number

Cf. A286838, A286839, A322087, A324077, A324082, A324083, A324084, A324085, A324086, A324087.

R:= select(t -> op([1,3,1],t)=11, [padic:-rootp(x^4-3, 13,101)]):
op([1,1,3],R); # Robert Israel, Sep 08 2019

a(n) = lift(-sqrtn(3+O(13^(n+1)), 4) * sqrt(-1+O(13^(n+1))))\13^n

Equals A324086*A286839 = A324087*A286838.
a(n) = (A324084(n+1) - A324084(n))/13^n.
For n > 0, a(n) = 12 - A324085(n).

A324085 Digits of one of the four 3-adic integers 3^(1/4) that is congruent to 2 mod 13.

Original entry on oeis.org

2, 2, 7, 1, 12, 6, 12, 4, 8, 6, 1, 10, 4, 6, 7, 8, 10, 1, 12, 9, 9, 7, 0, 12, 3, 6, 4, 5, 11, 12, 3, 11, 9, 5, 8, 4, 4, 2, 7, 4, 11, 8, 4, 10, 1, 0, 2, 1, 4, 3, 11, 7, 3, 6, 3, 2, 6, 7, 3, 6, 1, 0, 3, 0, 11, 8, 11, 6, 11, 0, 3, 5, 4, 7, 9, 10, 12, 6, 11, 5, 1

Offset: 0

Jianing Song, Sep 01 2019

One of the two square roots of A322087, where an A-number represents a 13-adic number. The other square root is A324153.

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.

			The unique number k in [1, 13^3] and congruent to 2 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 1211 = (722)_13, so the first three terms are 2, 7 and 7.

Wikipedia, p-adic number

Cf. A286838, A286839, A322087, A324077, A324082, A324083, A324084, A324086, A324087, A324153.

a(n) = lift(sqrtn(3+O(13^(n+1)), 4) * sqrt(-1+O(13^(n+1))))\13^n

Equals A324086*A286838 = A324087*A286839.
a(n) = (A324077(n+1) - A324077(n))/13^n.
For n > 0, a(n) = 12 - A324153(n).

A324086 Digits of one of the four 3-adic integers 3^(1/4) that is congruent to 3 mod 13.

Original entry on oeis.org

3, 5, 3, 6, 5, 12, 10, 2, 12, 12, 8, 12, 11, 7, 0, 2, 5, 11, 11, 3, 5, 11, 5, 4, 12, 12, 3, 2, 7, 7, 12, 11, 8, 5, 12, 3, 5, 8, 6, 12, 9, 4, 0, 5, 5, 12, 1, 9, 1, 9, 11, 7, 4, 0, 3, 9, 0, 12, 6, 6, 1, 8, 4, 9, 5, 6, 9, 5, 7, 10, 1, 3, 3, 8, 5, 11, 8, 2, 0, 1, 12

Offset: 0

Jianing Song, Sep 01 2019

One of the two square roots of A322088, where an A-number represents a 13-adic number. The other square root is A324087.

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.

			The unique number k in [1, 13^3] and congruent to 3 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 575 = (353)_13, so the first three terms are 3, 5 and 3.

Wikipedia, p-adic number

Cf. A286838, A286839, A322087, A324077, A324082, A324083, A324084, A324085, A324087, A324153.

a(n) = lift(sqrtn(3+O(13^(n+1)), 4))\13^n

Equals A324085*A286839 = A324153*A286838.
a(n) = (A324082(n+1) - A324082(n))/13^n.
For n > 0, a(n) = 12 - A324087(n).

A324087 Digits of one of the four 3-adic integers 3^(1/4) that is congruent to 10 mod 13.

Original entry on oeis.org

10, 7, 9, 6, 7, 0, 2, 10, 0, 0, 4, 0, 1, 5, 12, 10, 7, 1, 1, 9, 7, 1, 7, 8, 0, 0, 9, 10, 5, 5, 0, 1, 4, 7, 0, 9, 7, 4, 6, 0, 3, 8, 12, 7, 7, 0, 11, 3, 11, 3, 1, 5, 8, 12, 9, 3, 12, 0, 6, 6, 11, 4, 8, 3, 7, 6, 3, 7, 5, 2, 11, 9, 9, 4, 7, 1, 4, 10, 12, 11, 0

Offset: 0

Jianing Song, Sep 01 2019

One of the two square roots of A322088, where an A-number represents a 13-adic number. The other square root is A324086.

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.

			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 = (97A)_13, so the first three terms are 10, 7 and 9.

Wikipedia, p-adic number

Cf. A286838, A286839, A322087, A324077, A324082, A324083, A324084, A324085, A324086, A324153.

a(n) = lift(-sqrtn(3+O(13^(n+1)), 4))\13^n

Equals A324085*A286838 = A324153*A286839.
a(n) = (A324083(n+1) - A324083(n))/13^n.
For n > 0, a(n) = 12 - A324086(n).

A322088 Digits of one of the two 13-adic integers sqrt(3).

Original entry on oeis.org

9, 4, 6, 4, 0, 10, 11, 3, 3, 2, 11, 6, 8, 2, 6, 1, 1, 3, 7, 7, 12, 7, 10, 7, 4, 12, 4, 5, 9, 7, 9, 0, 12, 9, 2, 9, 7, 4, 11, 0, 1, 4, 5, 12, 9, 11, 8, 3, 3, 3, 11, 2, 6, 0, 10, 5, 9, 7, 11, 6, 0, 11, 11, 0, 2, 7, 6, 1, 5, 4, 0, 2, 11, 9, 7, 7, 7, 5, 1, 11, 7

Offset: 0

Jianing Song, Nov 26 2018

This square root of 3 in the 13-adic field ends with digit 9. The other, A322087, ends with digit 4.

			...10B47929C097954C47A7C773116286B233BA04649.

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. A286838, A286839, A322086, A322087, A322091, A322092.

a(n) = truncate(-sqrt(3+O(13^(n+1))))\13^n

a(n) = (A322086(n+1) - A322086(n))/13^n.
For n > 0, a(n) = 12 - A322087(n).
Equals A286838*A322092 = A286839*A322091.
This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {2*T(13^n,9/2)}, where T(n,x) denotes the n-th Chebyshev polynomial. - Peter Bala, Dec 04 2022

A322085 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 4 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 4, 108, 1122, 18698, 361430, 1104016, 5930825, 570667478, 7912243967, 113957237697, 251815729546, 11004778093768, 104197118583692, 3132948184506222, 26757206498701956, 589802029653700283, 7909384730668678534, 85763128005100719931, 648040162764887685576

Offset: 0

Jianing Song, Nov 26 2018

nonn

For n > 0, a(n) is the unique solution to x^2 == 3 (mod 13^n) in the range [0, 13^n - 1] and congruent to 4 modulo 13.

A322086 is the approximation (congruent to 9 mod 13) of another square root of 3 over the 13-adic field.

			4^2 = 16 = 1*13 + 3.
108^2 = 11664 = 69*13^2 + 3.
1122^2 = 1258884 = 573*13^3 + 3.

Robert Israel, Table of n, a(n) for n = 0..896
Wikipedia, p-adic number

Cf. A286840, A286841, A322086, A322087, A322089, A322090.

S:= map(t -> op([1,3],t),[padic:-evalp(RootOf(x^2-3,x),13,30)]):
S4:= op(select(t -> t[1]=4, S)):
seq(add(S4[i]*13^(i-1),i=1..n-1),n=1..31); # Robert Israel, Jun 13 2019

PARI
```
a(n) = truncate(sqrt(3+O(13^n)))
```

For n > 0, a(n) = 13^n - A322086(n).
a(n) = Sum_{i=0..n-1} A322087(i)*13^i.
a(n) = A286840(n)*A322089(n) mod 13^n = A286841(n)*A322090(n) mod 13^n.

A309989 Digits of one of the two 17-adic integers sqrt(-1).

Original entry on oeis.org

4, 2, 10, 5, 12, 16, 12, 8, 13, 3, 14, 0, 6, 1, 0, 15, 1, 8, 14, 5, 7, 16, 14, 1, 5, 13, 9, 6, 5, 12, 16, 15, 9, 16, 14, 12, 16, 1, 3, 6, 4, 10, 15, 5, 16, 12, 2, 1, 5, 4, 0, 15, 2, 11, 14, 9, 5, 1, 11, 16, 15, 7, 5, 6, 14, 3, 12, 0, 0, 11, 12, 13, 9, 5, 4, 16, 13

Offset: 0

Jianing Song, Aug 26 2019

This square root of -1 in the 17-adic field ends with digit 4. The other, A309990, ends with digit 13 (D when written as a 17-adic number).

			The solution to x^2 == -1 (mod 17^4) such that x == 4 (mod 17) is x == 27493 (mod 17^4), and 27493 is written as 5A24 in heptadecimal, so the first four terms are 4, 2, 10 and 5.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number

Cf. A286877, A286878.
Digits of p-adic square roots:
A318962, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A051277, A290558 (7-adic, sqrt(2));
A321074, A321075 (11-adic, sqrt(3));
A321078, A321079 (11-adic, sqrt(5));
A322091, A322092 (13-adic, sqrt(-3));
A286838, A286839 (13-adic, sqrt(-1));
A322087, A322088 (13-adic, sqrt(3));
this sequence, A309990 (17-adic, sqrt(-1)).

a(n) = truncate(sqrt(-1+O(17^(n+1))))\17^n

a(n) = (A286877(n+1) - A286877(n))/17^n.

For n > 0, a(n) = 16 - A309990(n).

A309990 Digits of one of the two 17-adic integers sqrt(-1).

Original entry on oeis.org

13, 14, 6, 11, 4, 0, 4, 8, 3, 13, 2, 16, 10, 15, 16, 1, 15, 8, 2, 11, 9, 0, 2, 15, 11, 3, 7, 10, 11, 4, 0, 1, 7, 0, 2, 4, 0, 15, 13, 10, 12, 6, 1, 11, 0, 4, 14, 15, 11, 12, 16, 1, 14, 5, 2, 7, 11, 15, 5, 0, 1, 9, 11, 10, 2, 13, 4, 16, 16, 5, 4, 3, 7, 11, 12, 0

Offset: 0

Jianing Song, Aug 26 2019

This square root of -1 in the 17-adic field ends with digit 13 (D when written as a 17-adic number). The other, A309989, ends with digit 4.

			The solution to x^2 == -1 (mod 17^4) such that x == 13 (mod 17) is x == 56028 (mod 17^4), and 56028 is written as B6ED in heptadecimal, so the first four terms are 13, 14, 6 and 11.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number

Cf. A286877, A286878.
Digits of p-adic square roots:
A318962, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A051277, A290558 (7-adic, sqrt(2));
A321074, A321075 (11-adic, sqrt(3));
A321078, A321079 (11-adic, sqrt(5));
A322091, A322092 (13-adic, sqrt(-3));
A286838, A286839 (13-adic, sqrt(-1));
A322087, A322088 (13-adic, sqrt(3));
A309989, this sequence (17-adic, sqrt(-1)).

a(n) = truncate(-sqrt(-1+O(17^(n+1))))\17^n

a(n) = (A286878(n+1) - A286878(n))/17^n.

For n > 0, a(n) = 16 - A309989(n).

A322091 Digits of one of the two 13-adic integers sqrt(-3).

Original entry on oeis.org

6, 3, 12, 6, 10, 7, 4, 4, 9, 8, 9, 2, 8, 5, 12, 3, 5, 4, 0, 6, 5, 1, 2, 6, 5, 9, 4, 9, 1, 1, 4, 6, 11, 3, 1, 12, 5, 2, 2, 6, 3, 11, 11, 8, 4, 5, 10, 10, 7, 9, 5, 7, 7, 7, 8, 0, 1, 0, 7, 7, 0, 9, 12, 10, 8, 1, 6, 1, 2, 10, 2, 9, 7, 2, 1, 10, 11, 4, 3, 5, 6

Offset: 0

Jianing Song, Nov 26 2018

This square root of -3 in the 13-adic field ends with digit 6. The other, A322092, ends with digit 7.

			...36225C13B64119495621560453C582989447A6C36.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.
Wikipedia, p-adic number

Cf. A114525, A286838, A286839, A322087, A322088, A322089, A322092.

a(n) = truncate(sqrt(-3+O(13^(n+1))))\13^n

a(n) = (A322089(n+1) - A322089(n))/13^n.
For n > 0, a(n) = 12 - A322092(n).
Equals A286838*A322088 = A286839*A322087.
This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {L(13^n,6)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022

A322092 Digits of one of the two 13-adic integers sqrt(-3).

Original entry on oeis.org

7, 9, 0, 6, 2, 5, 8, 8, 3, 4, 3, 10, 4, 7, 0, 9, 7, 8, 12, 6, 7, 11, 10, 6, 7, 3, 8, 3, 11, 11, 8, 6, 1, 9, 11, 0, 7, 10, 10, 6, 9, 1, 1, 4, 8, 7, 2, 2, 5, 3, 7, 5, 5, 5, 4, 12, 11, 12, 5, 5, 12, 3, 0, 2, 4, 11, 6, 11, 10, 2, 10, 3, 5, 10, 11, 2, 1, 8, 9, 7, 6

Offset: 0

Jianing Song, Nov 26 2018

This square root of -3 in the 13-adic field ends with digit 7. The other, A322091, ends with digit 6.

			...96AA70B9168BB38376AB76C879074A34388526097.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.
Wikipedia, p-adic number

Cf. A114525, A286838, A286839, A322087, A322088, A322090, A322091.

a(n) = truncate(-sqrt(-3+O(13^(n+1))))\13^n

a(n) = (A322090(n+1) - A322090(n))/13^n.
For n > 0, a(n) = 12 - A322091(n).
Equals A286838*A322087 = A286839*A322088.
This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {L(13^n,7)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022

cp's OEIS Frontend

A324153 Digits of one of the four 13-adic integers 3^(1/4) that is congruent to 11 mod 13.

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A324085 Digits of one of the four 3-adic integers 3^(1/4) that is congruent to 2 mod 13.

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A324086 Digits of one of the four 3-adic integers 3^(1/4) that is congruent to 3 mod 13.

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A324087 Digits of one of the four 3-adic integers 3^(1/4) that is congruent to 10 mod 13.

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A322088 Digits of one of the two 13-adic integers sqrt(3).

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A322085 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 4 (mod 13) case (except for n = 0).

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A309989 Digits of one of the two 17-adic integers sqrt(-1).

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