A286838 -id:A286838 - cp's OEIS frontend

A286840 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-1). Here the 5 (mod 13) case (except for n=0).

0, 5, 70, 239, 239, 143044, 1999509, 6826318, 6826318, 822557039, 85658552023, 1188526486815, 11941488851037, 291518510320809, 2108769149874327, 13920898306972194, 13920898306972194, 2675587335039691558, 63228498770709057089, 513050126578538629605

Offset: 0

Seiichi Manyama, Aug 01 2017

Seiichi Manyama, Table of n, a(n) for n = 0..897
Wikipedia, Hensel's Lemma.

The two successive approximations up to p^n for p-adic integer sqrt(-1): A048898 and A048899 (p=5), this sequence and A286841 (p=13), A286877 and A286878 (p=17).

Cf. A034944, A114525, A286838.

{0}~Join~Table[#&@@Select[PowerModList[-1, 1/2, 13^k], Mod[#, 13] == 5 &], {k, 20}]  (* Giorgos Kalogeropoulos, Oct 21 2022 *)

a(n) = truncate(sqrt(-1+O(13^n))); \\ Michel Marcus, Aug 04 2017

def A(k, m, n):
    ary=[0]
    a, mod = k, m
    for i in range(n):
          b=a%mod
          ary.append(b)
          a=b**m
          mod*=m
    return ary
def a286840(n):
    return A(5, 13, n)
print(a286840(100)) # Indranil Ghosh, Aug 03 2017, after Ruby

def A(k, m, n)
  ary = [0]
  a, mod = k, m
  n.times{
    b = a % mod
    ary << b
    a = b ** m
    mod *= m
  }
  ary
end
def A286840(n)
  A(5, 13, n)
end
p A286840(100)

a(0) = 0 and a(1) = 5, a(n) = a(n-1) + 9 * (a(n-1)^2 + 1) mod 13^n for n > 1.

a(n) == L(13^n,5) (mod 13^n) == ((5 + sqrt(29))/2)^(13^n) + ((5 - sqrt(29))/2)^(13^n) (mod 13^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 20 2022

A286839 Digits of one of the two 13-adic integers sqrt(-1) (digits 0, 1, ... , 9, 10, 11, 12 are used instead of 0, 1, ... , 9, A, B, C).

Original entry on oeis.org

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

Offset: 0

Seiichi Manyama, Aug 01 2017

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Digits of one of the two p-adic integers sqrt(-1): A210850 and A210851 (p=5), A286838 and this sequence (p=13).

def A(k, m, n)
  d_ary = []
  ary = [0]
  a, mod = k, m
  (n + 1).times{|i|
    b = a % mod
    d_ary << (b - ary[-1]) / m ** i
    ary << b
    a = b ** m
    mod *= m
  }
  d_ary
end
def A286839(n)
  A(8, 13, n)
end
p A286839(100)

a(n) = 12 - A286838(n) for n > 0.

Keyword "base" added by Jianing Song, Feb 17 2021

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

Original entry on oeis.org

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).

A322087 Digits of one of the two 13-adic integers sqrt(3).

Original entry on oeis.org

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

Offset: 0

Jianing Song, Nov 26 2018

This square root of 3 in the 13-adic field ends with digit 4. The other, A322088, ends with digit 9.

			...BC1853A30C35378085250559BB6A461A9912C8684.

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. A322085.
Digits of p-adic integers:
A321074, A321075 (11-adic, sqrt(3));
this sequence, A322088 (13-adic, sqrt(3));
A286838, A286839 (13-adic, sqrt(-1));
A322091, A322092 (13-adic, sqrt(-3)).

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

a(n) = (A322085(n+1) - A322085(n))/13^n.
For n > 0, a(n) = 12 - A322088(n).
Equals A286838*A322091 = A286839*A322092.
This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {2*T(13^n,2)}, where T(n,x) denotes the n-th Chebyshev polynomial. - Peter Bala, Dec 04 2022

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

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

cp's OEIS Frontend

A286840 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-1). Here the 5 (mod 13) case (except for n=0).

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A286839 Digits of one of the two 13-adic integers sqrt(-1) (digits 0, 1, ... , 9, 10, 11, 12 are used instead of 0, 1, ... , 9, A, B, C).

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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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A322087 Digits of one of the two 13-adic integers sqrt(3).

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