A048898 -id:A048898 - cp's OEIS frontend

A269590 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 4 mod 5 numbers (except for n=0).

0, 4, 14, 114, 364, 989, 13489, 13489, 169739, 560364, 2513489, 2513489, 2513489, 246654114, 3908763489, 22219310364, 52736888489, 52736888489, 3104494700989, 6919191966614

Offset: 0

Wolfdieter Lang, Mar 02 2016

The other approximation for the 5-adic integer sqrt(-4) with numbers 1 (mod 5) is given in A268922. See this also for more details and references.

For the digits of this 5-adic integer sqrt(-4), that is the scaled first differences, see A269592. See also A268922 for the 5-adic numbers u and -u written from the right to the left.

Seiichi Manyama, Table of n, a(n) for n = 0..1432

Cf. A048899, A048898, A269591, A268922, A269592.

with(padic): D2:=op(3,op([evalp(RootOf(x^2+4),5,20)][2])):
0,seq(sum('D2[k]*5^(k-1)','k'=1..n),n=1..20);
# alternative program
a := proc(n) option remember; if n = 1 then 4 else irem( a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1), 5^n) end if; end: seq(a(n), n = 1..20); # Peter Bala, Nov 14 2022

a(n) = if (n==0, 0, 5^n - truncate(sqrt(-4+O(5^(n))))); \\ Michel Marcus, Mar 07 2016

Recurrence for n >= 1: a(n) = modp( a(n-1) + 3*(a(n-1)^2 + 4), 5^n), n >= 2, with a(1) = 4. Here modp(a, m) is used to pick the representative of the residue class a modulo m from the smallest nonnegative complete residue system {0, 1, ... , m-1}.
a(n) = 5^n - A268922(n), n >= 1.
a(n) == Lucas(3*(5^n)) (mod 5^n). - Peter Bala, Nov 14 2022

A286877 One of the two successive approximations up to 17^n for 17-adic integer sqrt(-1). Here the 4 (mod 17) case (except for n=0).

Original entry on oeis.org

0, 4, 38, 2928, 27493, 1029745, 23747457, 313398285, 3596107669, 94280954402, 450044583893, 28673959190179, 28673959190179, 3524407382568745, 13428985415474682, 13428985415474682, 42949774758062711577, 91610966633729580058, 6709533061724423693474

Offset: 0

Seiichi Manyama, Aug 02 2017

x = ...GC5A24,

x^2 = ...GGGGGG = -1.

			a(1) = (   4)_17 = 4,
a(2) = (  24)_17 = 38,
a(3) = ( A24)_17 = 2928,
a(4) = (5A24)_17 = 27493.

Seiichi Manyama, Table of n, a(n) for n = 0..813
Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.
Wikipedia, Hensel's Lemma.

The two successive approximations up to p^n for p-adic integer sqrt(-1): A048898 and A048899 (p=5), A286840 and A286841 (p=13), this sequence and A286878 (p=17).

a(n) = truncate(sqrt(-1+O(17^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 a286877(n):
    return A(4, 17, n)
print(a286877(100)) # Indranil Ghosh, Aug 03 2017

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 A286877(n)
  A(4, 17, n)
end
p A286877(100)

a(0) = 0 and a(1) = 4, a(n) = a(n-1) + 2 * (a(n-1)^2 + 1) mod 17^n for n > 1.

a(n) == L(17^n,4) (mod 17^n) == (2 + sqrt(5))^(17^n) + (2 - sqrt(5))^(17^n) (mod 17^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Dec 02 2022

A286878 One of the two successive approximations up to 17^n for 17-adic integer sqrt(-1). Here the 13 (mod 17) case (except for n=0).

Original entry on oeis.org

0, 13, 251, 1985, 56028, 390112, 390112, 96940388, 3379649772, 24306922095, 1565949316556, 5597937117454, 553948278039582, 6380170650337192, 154948841143926247, 2848994066094341111, 5711417117604156904, 735629295252607184119, 7353551390343301297535

Offset: 0

Seiichi Manyama, Aug 02 2017

nonn

x = ...04B6ED,

x^2 = ...GGGGGG = -1.

			a(1) = (   D)_17 = 13,
a(2) = (  ED)_17 = 251,
a(3) = ( 6ED)_17 = 1985,
a(4) = (B6ED)_17 = 56028.

Seiichi Manyama, Table of n, a(n) for n = 0..812
Wikipedia, Hensel's Lemma.

The two successive approximations up to p^n for p-adic integer sqrt(-1): A048898 and A048899 (p=5), A286840 and A286841 (p=13), A286877 and this sequence (p=17).

a(n) = if (n, 17^n-truncate(sqrt(-1+O(17^n))), 0); \\ 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 a286878(n): return A(13, 17, n)
print(a286878(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 A286878(n)
  A(13, 17, n)
end
p A286878(100)

If n > 0, a(n) = 17^n - A286877(n).

a(0) = 0 and a(1) = 13, a(n) = a(n-1) + 15 * (a(n-1)^2 + 1) mod 17^n for n > 1.

A309444 The successive approximations up to 5^n for 5-adic integer 4^(1/3).

Original entry on oeis.org

0, 4, 9, 59, 559, 3059, 12434, 59309, 371809, 371809, 8184309, 27715559, 76543684, 320684309, 1541387434, 25955449934, 86990606184, 392166387434, 2680984746809, 14125076543684, 52272049199934, 338374344121809, 2245722976934309, 7014094558965559, 42776881424199934

Offset: 0

Seiichi Manyama, Aug 03 2019

nonn

			a(1) = (   4)_5 = 4,
a(2) = (  14)_5 = 9,
a(3) = ( 214)_5 = 59,
a(4) = (4214)_5 = 559.

Cf. A309443.
Expansions of p-adic integers:
A268922, A269590 (5-adic, sqrt(-4));
A048898, A048899 (5-adic, sqrt(-1));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3)).

PARI
```
{a(n) = truncate((4+O(5^n))^(1/3))}
```

a(0) = 0 and a(1) = 4, a(n) = a(n-1) + 3 * (a(n-1)^3 - 4) mod 5^n for n > 1.

A325485 One of the four successive approximations up to 5^n for the 5-adic integer 6^(1/4). This is the 2 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 2, 22, 22, 397, 397, 6647, 6647, 319147, 319147, 6178522, 6178522, 103834772, 592116022, 3033522272, 9137037897, 70172194147, 222760084772, 3274517897272, 3274517897272, 60494976881647, 441964703444147, 1395639019850397, 3779824810866022, 51463540631178522

Offset: 0

Jianing Song, Sep 07 2019

nonn

For n > 0, a(n) is the unique number k in [1, 5^n] and congruent to 2 mod 5 such that k^4 - 6 is divisible by 5^n.

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, 5^2] and congruent to 2 modulo 5 such that k^4 - 6 is divisible by 5^2 is k = 22, so a(2) = 22.
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 also k = 22, so a(3) is also 22.

Wikipedia, p-adic number

Cf. A048898, A048899, A324024, A325489, A325490, A325491, A325492.
Approximations of p-adic fourth-power roots:
A325484, this sequence, A325486, A325487 (5-adic, 6^(1/4));
A324077, A324082, A324083, A324084 (13-adic, 3^(1/4)).

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

a(n) = A325484(n)*A048898(n) mod 13^n = A325485(n)*A048899(n) mod 13^n.
For n > 0, a(n) = 5^n - A325486(n).
a(n)^2 == A324024(n) (mod 5^n).

A324023 One of the two successive approximations up to 5^n for 5-adic integer sqrt(6). This is the 1 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 1, 16, 16, 516, 1766, 4891, 36141, 270516, 661141, 6520516, 35817391, 35817391, 768239266, 4430348641, 16637379891, 108190114266, 413365895516, 1939244801766, 9568639333016, 85862584645516, 371964879567391, 1802476354176766, 4186662145192391, 51870377965504891

Offset: 0

Jianing Song, Sep 07 2019

nonn

For n > 0, a(n) is the unique solution to x^2 == 6 (mod 5^n) in the range [0, 5^n - 1] and congruent to 1 modulo 5.

A324024 is the approximation (congruent to 4 mod 5) of another square root of 6 over the 5-adic field.

			16^2 = 256 = 10*5^2 + 6 = 2*5^3 + 6;
516^2 = 266256 = 426*5^4 + 6;
1766^2 = 3118756 = 998*5^5 + 6.

Wikipedia, p-adic number

Cf. A048898, A048899, A324025, A324026.
Approximations of 5-adic square roots:
A324027, A324028 (sqrt(-6));
A268922, A269590 (sqrt(-4));
A048898, A048899 (sqrt(-1));
this sequence, A324024 (sqrt(6)).

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

For n > 0, a(n) = 5^n - A324024(n).

a(n) = A048898(n)*A324028(n) mod 5^n = A048899(n)*A324027(n) mod 5^n.

A324024 One of the two successive approximations up to 5^n for 5-adic integer sqrt(6). This is the 4 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 4, 9, 109, 109, 1359, 10734, 41984, 120109, 1291984, 3245109, 13010734, 208323234, 452463859, 1673166984, 13880198234, 44397776359, 349573557609, 1875452463859, 9504846995109, 9504846995109, 104872278635734, 581709436838859, 7734266809885734, 7734266809885734

Offset: 0

Jianing Song, Sep 07 2019

nonn

For n > 0, a(n) is the unique solution to x^2 == 6 (mod 5^n) in the range [0, 5^n - 1] and congruent to 1 modulo 5.

A324023 is the approximation (congruent to 4 mod 5) of another square root of 6 over the 5-adic field.

			9^2 = 81 = 3*5^2 + 6;
109^2 = 11881 = 95*5^3 + 6 = 19*5^4 + 6;
1359^2 = 1846881 = 591*5^5 + 6.

Wikipedia, p-adic number

Cf. A048898, A048899, A324025, A324026.
Approximations of 5-adic square roots:
A324027, A324028 (sqrt(-6));
A268922, A269590 (sqrt(-4));
A048898, A048899 (sqrt(-1));
A324023, this sequence (sqrt(6)).

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

For n > 0, a(n) = 5^n - A324023(n).

a(n) = A048898(n)*A324027(n) mod 5^n = A048899(n)*A324028(n) mod 5^n.

A325484 One of the four successive approximations up to 5^n for the 5-adic integer 6^(1/4). This is the 1 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 1, 21, 121, 246, 2121, 5246, 52121, 286496, 677121, 677121, 20208371, 117864621, 606145871, 3047552121, 3047552121, 94600286496, 704951848996, 2993770208371, 2993770208371, 79287715520871, 270022578802121, 746859737005246, 5515231319036496, 29357089229192746

Offset: 0

Jianing Song, Sep 07 2019

nonn

For n > 0, a(n) is the unique number k in [1, 5^n] and congruent to 1 mod 5 such that k^4 - 6 is divisible by 5^n.

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, 5^2] and congruent to 1 modulo 5 such that k^4 - 6 is divisible by 5^2 is k = 21, so a(2) = 21.
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, so a(3) = 121.

Wikipedia, p-adic number

Cf. A048898, A048899, A324023, A325489, A325490, A325491, A325492.
Approximations of p-adic fourth-power roots:
this sequence, A325485, A325486, A325487 (5-adic, 6^(1/4));
A324077, A324082, A324083, A324084 (13-adic, 3^(1/4)).

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

a(n) = A325485(n)*A048899(n) mod 5^n = A325486(n)*A048898(n) mod 5^n.
For n > 0, a(n) = 5^n - A325487(n).
a(n)^2 == A324023(n) (mod 5^n).

A325486 One of the four successive approximations up to 5^n for the 5-adic integer 6^(1/4). This is the 3 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 3, 3, 103, 228, 2728, 8978, 71478, 71478, 1633978, 3587103, 42649603, 140305853, 628587103, 3069993353, 21380540228, 82415696478, 540179368353, 540179368353, 15798968430853, 34872454758978, 34872454758978, 988546771165228, 8141104144212103, 8141104144212103

Offset: 0

Jianing Song, Sep 07 2019

nonn

For n > 0, a(n) is the unique number k in [1, 5^n] and congruent to 3 mod 5 such that k^4 - 6 is divisible by 5^n.

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, 5^2] and congruent to 3 modulo 5 such that k^4 - 6 is divisible by 5^2 is k = 3, so a(2) = 3.
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, so a(3) = 103.

Wikipedia, p-adic number

Cf. A048898, A048899, A324024, A325489, A325490, A325491, A325492.
Approximations of p-adic fourth-power roots:
A325484, A325485, this sequence, A325487 (5-adic, 6^(1/4));
A324077, A324082, A324083, A324084 (13-adic, 3^(1/4)).

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

a(n) = A325484(n)*A048899(n) mod 13^n = A325485(n)*A048898(n) mod 13^n.
For n > 0, a(n) = 5^n - A325485(n).
a(n)^2 == A324024(n) (mod 5^n).

A325487 One of the four successive approximations up to 13^n for the 13-adic integer 6^(1/4). This is the 4 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 4, 4, 4, 379, 1004, 10379, 26004, 104129, 1276004, 9088504, 28619754, 126276004, 614557254, 3055963504, 27470026004, 57987604129, 57987604129, 820927057254, 16079716119754, 16079716119754, 206814579401004, 1637326054010379, 6405697636041629, 30247555546197879

Offset: 0

Jianing Song, Sep 07 2019

nonn

For n > 0, a(n) is the unique number k in [1, 5^n] and congruent to 4 mod 5 such that k^4 - 6 is divisible by 5^n.

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, 5^2] and congruent to 4 modulo 5 such that k^4 - 6 is divisible by 5^2 is k = 4, so a(2) = 4.
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 also k = 4, so a(3) is also 4.

Wikipedia, p-adic number

Cf. A048898, A048899, A324023, A325489, A325490, A325491, A325492.
Approximations of p-adic fourth-power roots:
A325484, A325485, A325486, this sequence (5-adic, 6^(1/4));
A324077, A324082, A324083, A324084 (13-adic, 3^(1/4)).

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

a(n) = A325485(n)*A048898(n) mod 5^n = A325486(n)*A048899(n) mod 5^n.
For n > 0, a(n) = 5^n - A325484(n).
a(n)^2 == A324023(n) (mod 5^n).

cp's OEIS Frontend

A269590 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 4 mod 5 numbers (except for n=0).

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A286877 One of the two successive approximations up to 17^n for 17-adic integer sqrt(-1). Here the 4 (mod 17) case (except for n=0).

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A286878 One of the two successive approximations up to 17^n for 17-adic integer sqrt(-1). Here the 13 (mod 17) case (except for n=0).

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A309444 The successive approximations up to 5^n for 5-adic integer 4^(1/3).

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A325485 One of the four successive approximations up to 5^n for the 5-adic integer 6^(1/4). This is the 2 (mod 5) case (except for n = 0).

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A324023 One of the two successive approximations up to 5^n for 5-adic integer sqrt(6). This is the 1 (mod 5) case (except for n = 0).

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A324024 One of the two successive approximations up to 5^n for 5-adic integer sqrt(6). This is the 4 (mod 5) case (except for n = 0).

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