A210851 -id:A210851 - cp's OEIS frontend

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

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).

A212152 Digits of one of the three 7-adic integers (-1)^(1/3).

Original entry on oeis.org

3, 4, 6, 3, 0, 2, 6, 2, 4, 3, 4, 4, 5, 2, 1, 2, 1, 4, 6, 1, 1, 3, 5, 0, 2, 3, 4, 1, 3, 4, 3, 5, 6, 6, 2, 2, 2, 0, 2, 4, 0, 6, 6, 1, 5, 4, 1, 2, 3, 4, 1, 3, 4, 0, 3, 3, 2, 4, 4, 4, 5, 1, 0, 4, 0, 2, 0, 3, 1, 0, 2, 6, 1, 5, 2, 5, 5, 6, 0, 6, 2, 4, 4, 2, 1, 6, 3, 4, 5, 5, 1, 0, 4, 2, 4, 4, 5, 5, 1, 3

Offset: 0

Wolfdieter Lang, May 02 2012

See A210852 for comments and an approximation to this 7-adic number, called there u.  See also A048898 for references on p-adic numbers.
a(n), n>=1, is the (unique) solution of the linear congruence 3 * b(n)^2 * a(n) + c(n) == 0 (mod 7), with  b(n):=A210852(n) and c(n):=A210853(n). a(0) = 3, one of the three solutions of x^3+1 == 0 (mod 7).
Since b(n) == 3 (mod 7), a(n) == c(n) (mod 7) for n>0. - Álvar Ibeas, Feb 20 2017
With a(0) = 2, this is the digits of one of the three cube root of 1, the one that is congruent to 2 modulo 7. - Jianing Song, Aug 26 2022

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A210852 (approximations of (-1)^(1/3)), A212155 (digits of another cube root of -1), 6*A000012 (digits of -1).

Cf. A210850, A210851 (digits of the 5-adic integers sqrt(-1)); A319297, A319305, A319555 (digits of the 7-adic integers 6^(1/3)).

op([1,1,3],select(t -> padic:-ratvaluep(t,1)=3, [padic:-rootp(x^3+1,7,100)])); # Robert Israel, Mar 27 2018

Join[{3}, MapIndexed[#/7^#2[[1]] &, Differences[FoldList[PowerMod[#, 7, 7^#2] &, 3, Range[2, 100]]]]] (* Paolo Xausa, Jan 14 2025 *)

a(n) = (b(n+1) - b(n))/7^n, n>=1, with b(n):=A210852(n), defined by a recurrence given there. One also finds a Maple program for b(n) there. a(0)=3.

A212155 Digits of one of the three 7-adic integers (-1)^(1/3).

Original entry on oeis.org

5, 2, 0, 3, 6, 4, 0, 4, 2, 3, 2, 2, 1, 4, 5, 4, 5, 2, 0, 5, 5, 3, 1, 6, 4, 3, 2, 5, 3, 2, 3, 1, 0, 0, 4, 4, 4, 6, 4, 2, 6, 0, 0, 5, 1, 2, 5, 4, 3, 2, 5, 3, 2, 6, 3, 3, 4, 2, 2, 2, 1, 5, 6, 2, 6, 4, 6, 3, 5, 6, 4, 0, 5, 1, 4, 1, 1, 0, 6, 0, 4, 2, 2, 4, 5, 0, 3, 2, 1, 1, 5, 6, 2, 4, 2, 2, 1, 1, 5, 3

Offset: 0

Wolfdieter Lang, May 02 2012

See A210853 for comments and an approximation to this 7-adic number, called there v.  See also A048898 for references on p-adic numbers.
a(n), n>=1, is the (unique) solution of the linear congruence 3 * b(n)^2 * a(n) + c(n) == 0 (mod 7), with  b(n):=A212153(n) and c(n):=A212154(n). a(0) = 5, one of the three solutions of X^3+1 == 0 (mod 7).
Since b(n) == 5 (mod 7), a(n) == 4 * c(n) (mod 7) for n>0. - Álvar Ibeas, Feb 20 2017
With a(0) = 4, this is the digits of one of the three cube root of 1, the one that is congruent to 4 modulo 7. - Jianing Song, Aug 26 2022

Paolo Xausa, Table of n, a(n) for n = 0..10000

Cf. A212153 (approximations of (-1)^(1/3)), A212152 (digits of another cube root of -1), 6*A000012 (digits of -1).

Cf. A210850, A210851 (digits of the 5-adic integers sqrt(-1)); A319297, A319305, A319555 (digits of the 7-adic integers 6^(1/3)).

Join[{5}, MapIndexed[#/7^#2[[1]] &, Differences[FoldList[PowerMod[#, 7, 7^#2] &, 5, Range[2, 100]]]]] (* Paolo Xausa, Jan 14 2025 *)

a(n) = (b(n+1) - b(n))/7^n, n>=1, with b(n):=A212153(n), defined by a recurrence given there. One also finds there a Maple program for b(n). a(0)=5.

a(n) = 6 - A212152(n), for n>0. - Álvar Ibeas, Feb 21 2017

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).

cp's OEIS Frontend

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

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

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

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A212152 Digits of one of the three 7-adic integers (-1)^(1/3).

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A212155 Digits of one of the three 7-adic integers (-1)^(1/3).