A210850 -id:A210850 - cp's OEIS frontend

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

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).

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

A324029 Digits of one of the two 5-adic integers sqrt(-6) that is related to A324027.

Original entry on oeis.org

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

Offset: 0

Jianing Song, Sep 07 2019

This square root of -6 in the 5-adic field ends with digit 2. The other, A324030, ends with digit 3.

			The solution to x^2 == -6 (mod 5^4) such that x == 2 (mod 5) is x == 162 (mod 5^4), and 162 is written as 1122 in quinary, so the first four terms are 2, 2, 1 and 1.

Wikipedia, p-adic number

Cf. A324027, A324028.
Digits of 5-adic square roots:
this sequence, A324030 (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
A324025, A324026 (sqrt(6)).

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

a(n) = (A324027(n+1) - A324027(n))/5^n.
For n > 0, a(n) = 4 - A324030(n).
Equals A210850*A324026 = A210851*A324025, where each A-number represents a 5-adic number.

A324030 Digits of one of the two 5-adic integers sqrt(-6) that is related to A324028.

Original entry on oeis.org

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

Offset: 0

Jianing Song, Sep 07 2019

This square root of -6 in the 5-adic field ends with digit 3. The other, A324029, ends with digit 2.

			The solution to x^2 == -6 (mod 5^4) such that x == 3 (mod 5) is x == 463 (mod 5^4), and 463 is written as 3323 in quinary, so the first four terms are 3, 2, 3 and 3.

Wikipedia, p-adic number

Cf. A324027, A324028.
Digits of 5-adic square roots:
A324029, sequence (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
A324025, A324026 (sqrt(6)).

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

a(n) = (A324028(n+1) - A324028(n))/5^n.
For n > 0, a(n) = 4 - A324029(n).
Equals A210850*A324025 = A210851*A324026, where each A-number represents a 5-adic number.

A327305 Digits of one of the two 5-adic integers sqrt(-9) that is related to A327303.

Original entry on oeis.org

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

Offset: 0

Jianing Song, Sep 16 2019

This is the 5-adic solution to x^2 = -9 that ends in 4. A327304 gives the other solution that ends in 1.

			Equals ...1131142000414124220322114423220241100304.

Robert Israel, Table of n, a(n) for n = 0..10000
G. P. Michon, Introduction to p-adic integers, Numericana.

Cf. A327302, A327303.
Digits of 5-adic square roots:
A327304, this sequence (sqrt(-9));
A324029, A324030 (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
A324025, A324026 (sqrt(6)).

op([1,1,3], select(t -> padic:-ratvaluep(t,1)=4, [padic:-rootp(x^2+9,5,100)])); # Robert Israel, Aug 31 2020

a(n) = truncate(-sqrt(-9+O(5^(n+1))))\5^n

For n > 0, a(n) is the unique m in {0, 1, 2, 3, 4} such that (A327303(n) + m*5^n)^2 + 9 is divisible by 5^(n+1).
a(n) = (A327303(n+1) - A327303(n))/5^n.
For n > 0, a(n) = 4 - A327304(n).

A327304 Digits of one of the two 5-adic integers sqrt(-9) that is related to A327302.

Original entry on oeis.org

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

Offset: 0

Jianing Song, Sep 16 2019

This is the 5-adic solution to x^2 = -9 that ends in 1. A327305 gives the other solution that ends in 4.

			Equals ...3313302444030320224122330021224203344141.

G. P. Michon, Introduction to p-adic integers, Numericana.

Cf. A327302, A327303.
Digits of 5-adic square roots:
this sequence, A327305 (sqrt(-9));
A324029, A324030 (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
A324025, A324026 (sqrt(6)).

a(n) = truncate(-sqrt(-9+O(5^(n+1))))\5^n

For n > 0, a(n) is the unique m in {0, 1, 2, 3, 4} such that (A327302(n) + m*5^n)^2 + 9 is divisible by 5^(n+1).
a(n) = (A327302(n+1) - A327302(n))/5^n.
For n > 0, a(n) = 4 - A327305(n).

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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).

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

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

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

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

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A324029 Digits of one of the two 5-adic integers sqrt(-6) that is related to A324027.

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A324030 Digits of one of the two 5-adic integers sqrt(-6) that is related to A324028.

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