A210850 -id:A210850 - cp's OEIS frontend

A290803 One of the two successive approximations up to 7^n for the 7-adic integer sqrt(-3). These are the numbers congruent to 2 mod 7 (except for the initial 0).

0, 2, 37, 37, 2095, 14100, 47714, 165363, 988906, 29812911, 271934553, 1401835549, 9311142521, 36993716923, 133882727330, 2846775018726, 12341898038612, 145273620317016, 1308426190253051, 1308426190253051, 35505111746372480, 354674176936820484

Offset: 0

Seiichi Manyama, Aug 11 2017

x = ...256052,

x^2 = ...666664 = -3.

			a(1) =     2_7 = 2,
a(2) =    52_7 = 37,
a(3) =    52_7 = 37,
a(4) =  6052_7 = 2095,
a(5) = 56052_7 = 14100.

Seiichi Manyama, Table of n, a(n) for n = 0..1184
Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.
Wikipedia, Hensel's Lemma.

Cf. A114525, A290796, A290804.

a(n) = if (n, truncate(sqrt(-3+O(7^(n)))), 0)

a(0) = 0 and a(1) = 2, a(n) = a(n-1) + 5 * (a(n-1)^2 + 3) mod 7^n for n > 1.

a(n) = L(7^n,2) (mod 7^n) = ( (1 + sqrt(2))^(7^n) + (1 - sqrt(2))^(7^n) ) (mod 7^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 28 2022

A290804 One of the two successive approximations up to 7^n for the 7-adic integer sqrt(-3). These are the numbers congruent to 5 mod 7 (except for the initial 0).

Original entry on oeis.org

0, 5, 12, 306, 306, 2707, 69935, 658180, 4775895, 10540696, 10540696, 575491194, 4530144680, 59895293484, 544340345519, 1900786491217, 20891032530989, 87356893670191, 319987407657398, 10090468995120092, 44287154551239521, 203871687146463523

Offset: 0

Seiichi Manyama, Aug 11 2017

x = ...410615,

x^2 = ...666664 = -3.

			a(1) =     5_7 = 5,
a(2) =    15_7 = 12,
a(3) =   615_7 = 306,
a(4) =   615_7 = 306,
a(5) = 10615_7 = 2707.

Seiichi Manyama, Table of n, a(n) for n = 0..1183
Peter Bala, Using Lucas polynomials to find the p -adic square roots of -1, -2 and -3, Dec 2022.
Wikipedia, Hensel's Lemma.

Cf. A114525, A290797, A290803.

a(n) = if (n, 7^n - truncate(sqrt(-3+O(7^(n)))), 0)

a(0) = 0 and a(1) = 5, a(n) = a(n-1) + 2 * (a(n-1)^2 + 3) mod 7^n for n > 1.
If n > 0, a(n) = 7^n - A290803(n).
a(n) = L(7^n,5) (mod 7^n) = ( ((5 + sqrt(29))/2)^(7^n) + ((5 - sqrt(29))/2)^(7^n) ) (mod 7^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 28 2022

A325490 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 2 mod 5.

Original entry on oeis.org

2, 4, 0, 3, 0, 2, 0, 4, 0, 3, 0, 2, 2, 2, 1, 2, 1, 4, 0, 3, 4, 2, 1, 4, 1, 1, 2, 0, 0, 3, 0, 1, 1, 3, 1, 4, 4, 0, 2, 4, 0, 4, 1, 2, 0, 1, 2, 3, 2, 4, 2, 4, 1, 3, 0, 2, 1, 0, 3, 3, 3, 3, 0, 2, 2, 3, 1, 1, 4, 1, 1, 0, 1, 4, 0, 3, 3, 3, 0, 3, 0, 0, 4, 0, 3, 2, 3, 1

Offset: 0

Jianing Song, Sep 07 2019

One of the two square roots of A324026, where an A-number represents a 5-adic number. The other square root is A325491.

For k not divisible by 5, k is a fourth power in 5-adic field if and only if k == 1 (mod 5). If k is a fourth power in 5-adic field, then k has exactly 4 fourth-power roots.

			The unique number k in [1, 5^3] and congruent to 2 modulo 5 such that k^4 - 6 is divisible by 5^3 is k = 22 = (42)_5, so the first three terms are 2, 4 and 0.

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

Cf. A210850, A210851, A324026, A325484, A325485, A325486, A325487.
Digits of p-adic fourth-power roots:
A325489, this sequence, A325491, A325492 (5-adic, 6^(1/4));
A324085, A324086, A324087, A324153 (13-adic, 3^(1/4)).

S:= select(t -> op([1,3,1],t)=2, [padic:-rootp(_Z^4-6,5,100)]):
op([1,1,3],S); # Robert Israel, Mar 23 2023

a(n) = lift(sqrtn(6+O(5^(n+1)), 4) * sqrt(-1+O(5^(n+1))))\5^n

Equals A325489*A210850 = A325492*A210851.
a(n) = (A325485(n+1) - A325485(n))/13^n.
For n > 0, a(n) = 4 - A325491(n).

A318962 Digits of one of the two 2-adic integers sqrt(-7) that ends in 01.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Jianing Song, Sep 06 2018

Over the 2-adic integers there are 2 solutions to x^2 = -7, one ends in 01 and the other ends in 11. This sequence gives the former one. See A318960 for detailed information.

			...10110001110011100100110001100000010110101.

Jianing Song, Table of n, a(n) for n = 0..1000

Cf. A318960.
Digits of p-adic integers:
this sequence, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A212152, A212155 (7-adic, (1+sqrt(-3))/2);
A051277, A290558 (7-adic, sqrt(2));
A286838, A286839 (13-adic, sqrt(-1));
A309989, A309990 (17-adic, sqrt(-1)).
Also there are numerous sequences related to digits of 10-adic integers.

a(n) = truncate(-sqrt(-7+O(2^(n+2))))\2^n

a(0) = 1, a(1) = 0; for n >= 2, a(n) = 0 if A318960(n)^2 + 7 is divisible by 2^(n+2), otherwise 1.
a(n) = 1 - A318963(n) for n >= 1.
For n >= 2, a(n) = (A318960(n+1) - A318960(n))/2^n.

Corrected by Jianing Song, Aug 28 2019

A318963 Digits of one of the two 2-adic integers sqrt(-7) that ends in 11.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0

Offset: 0

Jianing Song, Sep 06 2018

Over the 2-adic integers there are 2 solutions to x^2 = -7, one ends in 01 and the other ends in 11. This sequence gives the latter one. See A318961 for detailed information.

			...01001110001100011011001110011111101001011.

Jianing Song, Table of n, a(n) for n = 0..1000

Cf. A318961.
Digits of p-adic integers:
A318962, this sequence (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A212152, A212155 (7-adic, (1+sqrt(-3))/2);
A051277, A290558 (7-adic, sqrt(2));
A286838, A286839 (13-adic, sqrt(-1));
A309989, A309990 (17-adic, sqrt(-1)).
Also there are numerous sequences related to digits of 10-adic integers.

a(n) = if(n==1, 1, truncate(sqrt(-7+O(2^(n+2))))\2^n)

a(0) = a(1) = 1; for n >= 2, a(n) = 0 if A318961(n)^2 + 7 is divisible by 2^(n+2), otherwise 1.
a(n) = 1 - A318962(n) for n >= 1.
For n >= 2, a(n) = (A318961(n+1) - A318961(n))/2^n.

Corrected by Jianing Song, Aug 28 2019

A324025 Digits of one of the two 5-adic integers sqrt(6) that is related to A324023.

Original entry on oeis.org

1, 3, 0, 4, 2, 1, 2, 3, 1, 3, 3, 0, 3, 3, 2, 3, 2, 2, 2, 4, 3, 3, 1, 4, 0, 1, 2, 0, 0, 0, 3, 3, 1, 4, 1, 0, 1, 2, 4, 1, 4, 1, 1, 0, 2, 4, 4, 3, 0, 2, 3, 4, 1, 1, 4, 3, 4, 2, 4, 2, 1, 1, 2, 4, 4, 3, 2, 3, 1, 1, 0, 1, 4, 2, 3, 4, 4, 4, 4, 0, 3, 3, 1, 2, 3, 2, 3, 1

Offset: 0

Jianing Song, Sep 07 2019

This square root of 6 in the 5-adic field ends with digit 1. The other, A324026, ends with digit 4.

			The solution to x^2 == 6 (mod 5^4) such that x == 1 (mod 5) is x == 516 (mod 5^4), and 516 is written as 4031 in quinary, so the first four terms are 1, 3, 0 and 4.

Wikipedia, p-adic number

Cf. A324023, A324024.
Digits of 5-adic square roots:
A324029, A324030 (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
this sequence, A324026 (sqrt(6)).

PARI
```
a(n) = truncate(sqrt(6+O(5^(n+1))))\5^n
```

a(n) = (A324023(n+1) - A324023(n))/5^n.
For n > 0, a(n) = 4 - A324026(n).
Equals A210850*A324030 = A210851*A324029, where each A-number represents a 5-adic number.

A324026 Digits of one of the two 5-adic integers sqrt(6) that is related to A324024.

Original entry on oeis.org

4, 1, 4, 0, 2, 3, 2, 1, 3, 1, 1, 4, 1, 1, 2, 1, 2, 2, 2, 0, 1, 1, 3, 0, 4, 3, 2, 4, 4, 4, 1, 1, 3, 0, 3, 4, 3, 2, 0, 3, 0, 3, 3, 4, 2, 0, 0, 1, 4, 2, 1, 0, 3, 3, 0, 1, 0, 2, 0, 2, 3, 3, 2, 0, 0, 1, 2, 1, 3, 3, 4, 3, 0, 2, 1, 0, 0, 0, 0, 4, 1, 1, 3, 2, 1, 2, 1, 3

Offset: 0

Jianing Song, Sep 07 2019

This square root of 6 in the 5-adic field ends with digit 4. The other, A324025, ends with digit 1.

			The solution to x^2 == 6 (mod 5^4) such that x == 4 (mod 5) is x == 109 (mod 5^4), and 109 is written as 414 in quinary, so the first four terms are 4, 1, 4 and 0.

Wikipedia, p-adic number

Cf. A324023, A324024.
Digits of 5-adic square roots:
A324029, A324030 (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
A324025, this sequence (sqrt(6)).

a(n) = truncate(-sqrt(6+O(5^(n+1))))\5^n

a(n) = (A324024(n+1) - A324024(n))/5^n.
For n > 0, a(n) = 4 - A324025(n).
Equals A210850*A324029 = A210851*A324030, where each A-number represents a 5-adic number.

A325489 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 1 mod 5.

Original entry on oeis.org

1, 4, 4, 1, 3, 1, 3, 3, 1, 0, 2, 2, 2, 2, 0, 3, 4, 3, 0, 4, 2, 1, 2, 2, 0, 1, 1, 2, 4, 2, 3, 4, 2, 1, 2, 3, 4, 3, 1, 0, 3, 2, 3, 4, 2, 3, 4, 4, 4, 2, 2, 2, 4, 1, 1, 0, 2, 1, 3, 3, 2, 0, 0, 1, 2, 4, 4, 1, 0, 4, 1, 0, 2, 4, 0, 2, 2, 0, 1, 3, 1, 1, 4, 3, 4, 1, 2, 2

Offset: 0

Jianing Song, Sep 07 2019

One of the two square roots of A324025, where an A-number represents a 5-adic number. The other square root is A325492.

For k not divisible by 5, k is a fourth power in 5-adic field if and only if k == 1 (mod 5). If k is a fourth power in 5-adic field, then k has exactly 4 fourth-power roots.

			The unique number k in [1, 5^3] and congruent to 1 modulo 5 such that k^4 - 6 is divisible by 5^3 is k = 121 = (441)_5, so the first three terms are 1, 4 and 4.

Wikipedia, p-adic number

Cf. A210850, A210851, A324025, A325484, A325485, A325486, A325487.
Digits of p-adic fourth-power roots:
this sequence, A325490, A325491, A325492 (5-adic, 6^(1/4));
A324085, A324086, A324087, A324153 (13-adic, 3^(1/4)).

PARI
```
a(n) = lift(sqrtn(6+O(5^(n+1)), 4))\5^n
```

Equals A325490*A210851 = A325491*A210850.
a(n) = (A325484(n+1) - A325484(n))/5^n.
For n > 0, a(n) = 4 - A325492(n).

A325491 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 3 mod 5.

Original entry on oeis.org

3, 0, 4, 1, 4, 2, 4, 0, 4, 1, 4, 2, 2, 2, 3, 2, 3, 0, 4, 1, 0, 2, 3, 0, 3, 3, 2, 4, 4, 1, 4, 3, 3, 1, 3, 0, 0, 4, 2, 0, 4, 0, 3, 2, 4, 3, 2, 1, 2, 0, 2, 0, 3, 1, 4, 2, 3, 4, 1, 1, 1, 1, 4, 2, 2, 1, 3, 3, 0, 3, 3, 4, 3, 0, 4, 1, 1, 1, 4, 1, 4, 4, 0, 4, 1, 2, 1, 3

Offset: 0

Jianing Song, Sep 07 2019

One of the two square roots of A324026, where an A-number represents a 5-adic number. The other square root is A325490.

For k not divisible by 5, k is a fourth power in 5-adic field if and only if k == 1 (mod 5). If k is a fourth power in 5-adic field, then k has exactly 4 fourth-power roots.

			The unique number k in [1, 5^3] and congruent to 3 modulo 5 such that k^4 - 6 is divisible by 5^3 is k = 103 = (403)_5, so the first three terms are 3, 0 and 4.

Wikipedia, p-adic number

Cf. A210850, A210851, A324026, A325484, A325485, A325486, A325487.
Digits of p-adic fourth-power roots:
A325489, A325490, this sequence, A325492 (5-adic, 6^(1/4));
A324085, A324086, A324087, A324153 (13-adic, 3^(1/4)).

a(n) = lift(-sqrtn(6+O(5^(n+1)), 4) * sqrt(-1+O(5^(n+1))))\5^n

Equals A325489*A210851 = A325492*A210850.
a(n) = (A325486(n+1) - A325486(n))/13^n.
For n > 0, a(n) = 4 - A325490(n).

A325492 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 4 mod 5.

Original entry on oeis.org

4, 0, 0, 3, 1, 3, 1, 1, 3, 4, 2, 2, 2, 2, 4, 1, 0, 1, 4, 0, 2, 3, 2, 2, 4, 3, 3, 2, 0, 2, 1, 0, 2, 3, 2, 1, 0, 1, 3, 4, 1, 2, 1, 0, 2, 1, 0, 0, 0, 2, 2, 2, 0, 3, 3, 4, 2, 3, 1, 1, 2, 4, 4, 3, 2, 0, 0, 3, 4, 0, 3, 4, 2, 0, 4, 2, 2, 4, 3, 1, 3, 3, 0, 1, 0, 3, 2, 2

Offset: 0

Jianing Song, Sep 07 2019

One of the two square roots of A324025, where an A-number represents a 5-adic number. The other square root is A325489.

For k not divisible by 5, k is a fourth power in 5-adic field if and only if k == 1 (mod 5). If k is a fourth power in 5-adic field, then k has exactly 4 fourth-power roots.

			The unique number k in [1, 5^3] and congruent to 4 modulo 5 such that k^4 - 6 is divisible by 5^3 is k = 4 = (4)_5, so the first three terms are 4, 0 and 0.

Wikipedia, p-adic number

Cf. A210850, A210851, A324025, A325484, A325485, A325486, A325487.
Digits of p-adic fourth-power roots:
A325489, A325490, A325491, this sequence (5-adic, 6^(1/4));
A324085, A324086, A324087, A324153 (13-adic, 3^(1/4)).

a(n) = lift(-sqrtn(6+O(5^(n+1)), 4))\5^n

Equals A325490*A210850 = A325491*A210851.
a(n) = (A325487(n+1) - A325487(n))/5^n.
For n > 0, a(n) = 4 - A325489(n).

cp's OEIS Frontend

A290803 One of the two successive approximations up to 7^n for the 7-adic integer sqrt(-3). These are the numbers congruent to 2 mod 7 (except for the initial 0).

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A290804 One of the two successive approximations up to 7^n for the 7-adic integer sqrt(-3). These are the numbers congruent to 5 mod 7 (except for the initial 0).

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A325490 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 2 mod 5.

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A318962 Digits of one of the two 2-adic integers sqrt(-7) that ends in 01.

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A318963 Digits of one of the two 2-adic integers sqrt(-7) that ends in 11.

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A324025 Digits of one of the two 5-adic integers sqrt(6) that is related to A324023.

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A324026 Digits of one of the two 5-adic integers sqrt(6) that is related to A324024.

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