A048899 -id:A048899 - 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.

A210848 a(n) = (A048898(n)^2 + 1)/5^n, n >= 0.

Original entry on oeis.org

1, 1, 2, 26, 53, 1354, 13562, 26858, 200965, 40193, 3859882, 13496122, 62298370, 12459674, 4106065226, 4044371993, 69072101242, 218014644394, 3137550252170, 627510050434, 66696011833378, 280704828874769, 2167389209973245, 433477841994649, 41870795375097221, 40277856145834642

Offset: 0

Wolfdieter Lang, Apr 28 2012

a(n) is an integer (nonnegative) because b(n):=A048898(n) satisfies b(n)^2 + 1 == 0 (mod 5^n), n>=0. The solution of this congruence for n>=1, which satisfies also b(n) == 2 (mod 5), is b(n) = 2^(5^(n-1)) (mod 5^n), but this is inconvenient for computing b(n) for large n. Instead one can use the b(n) recurrence which follows immediately, and this is given in the formula field below. To prove that the given b(n) formula solves the first congruence one can analyze the binomial expansion of (5 - 1)^(5^(n-1)) + 1 and show that it is 0 (mod 5^n) term by term. The second congruence reduces to b(n) == 2^(5^(n-1)) (mod 5) which follows for n>=1 by induction. Because b(n) = 5^n - A048899(n) one could also use the result A048899(n) == 3 (mod 5), once this has been proved.

The fact that X^2 + 1 == 0 (mod 5^n) has precisely two solutions for each n>=1, called x(n) and y(n), follows from the fact that X^2 + 1 == 0 (mod 5) has the two simple roots x(1) = 2 and y(1) = 3, and a theorem, given, e.g., in the Nagell reference as Theorem 50 on p. 87. From that same theorem, it also follows that one can choose all x(n) == 2 (mod 5) and all y(n) == 3 (mod 5).

			a(0) = 1/1 = 1.
a(3) = (57^2 + 1)/5^3 = 26 (b(3) = 7^5 (mod 5^3) = 57).

T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.

Paolo Xausa, Table of n, a(n) for n = 0..1000

Cf. A048898, A048899, A210849 (companion sequence).

b:=proc(n) option remember: if n=0 then 0 elif n=1 then 2
else modp(b(n-1)^5,5^n) fi: end proc:
[seq((b(n)^2+1)/5^n,n=0..29)];

Join[{1}, MapIndexed[(#^2 + 1)/5^#2[[1]] &, FoldList[PowerMod[#, 5, 5^#2] &, 2, Range[2, 25]]]] (* Paolo Xausa, Jan 14 2025 *)

a(n) = (A048898(n)^2 + 1)/5^n.

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).

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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A210848 a(n) = (A048898(n)^2 + 1)/5^n, 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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