A268922 -id:A268922 - cp's OEIS frontend

A269593 a(n) = (A268922(n)^2 + 4)/5^n, n >= 0.

4, 1, 5, 1, 109, 1460, 292, 53476, 124904, 993169, 5385572, 43930532, 239139524, 777233593, 789206948, 2256445369, 65340851012, 661111023620, 132222204724, 7745142596633, 10225443529556, 103321258570120, 20664251714024, 4562022446935993, 6246398287209928, 20888388201358465

Offset: 0

Wolfdieter Lang, Mar 02 2016

a(n) is an integer because b(n) = A268922(n) satisfies b(n)^2 + 4 == 0 (mod 5^n), n>=0.

See A268922 for details and references.

			a(0) = (0 + 4)/1 = 4.
a(4) = (261^2 + 4)/5^4 = 109.

Andrew Howroyd, Table of n, a(n) for n = 0..500

Cf. A268922, A269590, A269594 (companion sequence).

a(n) = ((truncate(sqrt(-4+O(5^(n)))))^2 + 4)/5^n; \\ Michel Marcus, Mar 07 2016

a(n) = (b(n)^2 + 1)/5^n, n>=0, with b(n) = A268922(n).

Terms a(21) and beyond from Andrew Howroyd, Mar 02 2020

A048898 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-1).

Original entry on oeis.org

0, 2, 7, 57, 182, 2057, 14557, 45807, 280182, 280182, 6139557, 25670807, 123327057, 123327057, 5006139557, 11109655182, 102662389557, 407838170807, 3459595983307, 3459595983307, 79753541295807, 365855836217682, 2273204469030182, 2273204469030182, 49956920289342682

Offset: 0

Michael Somos, Jul 26 1999

This is the root congruent to 2 mod 5.
Or, residues modulo 5^n of a 5-adic solution of x^2+1=0.
The radix-5 expansion of a(n) is obtained from the n rightmost digits in the expansion of the following pentadic integer:
  ...422331102414131141421404340423140223032431212 = u
The residues modulo 5^n of the other 5-adic solution of x^2+1=0 are given by A048899 which corresponds to the pentadic integer -u:
  ...022113342030313303023040104021304221412013233 = -u
The digits of u and -u are given in A210850 and A210851, respectively. - Wolfdieter Lang, May 02 2012
For approximations for p-adic square roots see also the W. Lang link under A268922. - Wolfdieter Lang, Apr 03 2016
From Jianing Song, Sep 06 2022: (Start)
For n > 0, a(n)-1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 1 modulo 5.
For n > 0, a(n)+1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 3 modulo 5. (End)

			a(0)=0 because 0 satisfies any equation in integers modulo 1.
a(1)=2 because 2 is one solution of x^2+1=0 modulo 5. (The other solution is 3, which gives rise to A048899.)
a(2)=7 because the equation (5y+a(1))^2+1=0 modulo 25 means that y is 1 modulo 5.

J. H. Conway, The Sensual Quadratic Form, p. 118, Mathematical Association of America, 1997, The Carus Mathematical Monographs, Number 26.
K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973, p. 35.

Seiichi Manyama, Table of n, a(n) for n = 0..1431 (terms 0..200 from Vincenzo Librandi)
Peter Bala, Using Lucas polynomials to find the p -adic square roots of -1, -2 and -3, Dec 2022.
G. P. Michon, On the witnesses of a composite integer, Numericana.
G. P. Michon, Introduction to p-adic integers, Numericana.

The two successive approximations up to p^n for p-adic integer sqrt(-1): this sequence and A048899 (p=5), A286840 and A286841 (p=13), A286877 and A286878 (p=17).
Cf. A000351 (powers of 5), A034939(n) = Min(a(n), A048899(n)).
Different from A034935.

[n le 2 select 2*(n-1) else Self(n-1)^5 mod 5^(n-1): n in [1..30]]; // Vincenzo Librandi, Feb 29 2016

a[0] = 0; a[1] = 2; a[n_] := a[n] = Mod[a[n-1]^5, 5^n]; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Nov 24 2011, after PARI *)
Join[{0}, RecurrenceTable[{a[1] == 2, a[n] == Mod[a[n-1]^5, 5^n]}, a, {n, 25}]] (* Vincenzo Librandi, Feb 29 2016 *)

PARI
```
{a(n) = if( n<2, 2, a(n-1)^5) % 5^n}
```

a(n) = lift(sqrt(-1 + O(5^n))); \\ Kevin Ryde, Dec 22 2020

If n>0, a(n) = 5^n - A048899(n).
From Wolfdieter Lang, Apr 28 2012: (Start)
Recurrence: a(n) = a(n-1)^5 (mod 5^n), a(1) = 2, n>=2. See the J.-F. Alcover Mathematica program and the PARI program below.
a(n) == 2^(5^(n-1)) (mod 5^n), n>=1.
a(n)*a(n-1) + 1 == 0 (mod 5^(n-1)), n>=1.
(a(n)^2 + 1)/5^n = A210848(n), n>=0.
(End)
Another recurrence: a(n) = modp(a(n-1) + a(n-1)^2 + 1, 5^n), n >= 2, a(1) = 2. Here modp(a, m) is the representative from {0, 1, ..., |m|-1} of the residue class a modulo m. Note that a(n) is in the residue class of a(n-1) modulo 5^(n-1) (see Hensel lifting). - Wolfdieter Lang, Feb 28 2016
a(n) == L(5^n,2) (mod 5^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 20 2022

Additional comments from Gerard P. Michon, Jul 15 2009
Edited by N. J. A. Sloane, Jul 25 2009
Name clarified by Wolfdieter Lang, Feb 19 2016

A269591 Digits of one of the two 5-adic integers sqrt(-4).

Original entry on oeis.org

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

Offset: 0

Wolfdieter Lang, Mar 02 2016

This is the scaled first difference sequence of A268922.
The digits of the other 5-adic integer sqrt(-4), are given in A269592. See also A268922 for the two 5-adic numbers sqrt(-4), called u and -u.
a(n) is the unique solution of the linear congruence 2*A268922(n)*a(n) + A269593(n) == 0 (mod 5), n>=1. Therefore only the values 0, 1, 2, 3 and 4 appear. See the Nagell reference given in A268922, eq. (6) on p. 86, adapted to this case. a(0) = 1 follows from the formula given below.

			a(4) = 3 because 2*261*3 + 109 = 1675 == 0 (mod 5).
a(4) = - 109*(2*261)^3 (mod 5) = -(-1)*(2*1)^3 (mod 5) = 8 (mod 5) = 3.
A268922(5) = 2136 = 1*5^0 + 2*5^1 + 0*5^2 + 2*5^3 + 3*5^4.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
BCMATH Congruence Programs, Finding a p-adic square root of a quadratic residue (mod p), p an odd prime.

Cf. A210850, A210851, A268922, A269592 (companion), A269593.

a(n) = truncate(sqrt(-4+O(5^(n+1))))\5^n; \\ Michel Marcus, Mar 04 2016

a(n) = (b(n+1) - b(n))/5^n, n >= 0, with b(n) = A268922(n), n >= 0.
a(n) = -A269593(n)*(2*A268922(n))^3 (mod 5), n >= 1. Solution of the linear congruence see, e.g., Nagell, Theorem 38 pp. 77-78.
A268922(n+1) = sum(a(k)*5^k, k=0..n), n >= 0.

A269592 Digits of one of the two 5-adic integers sqrt(-4). Here the ones related to A269590.

Original entry on oeis.org

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

Offset: 0

Wolfdieter Lang, Mar 02 2016

This is the scaled first difference sequence of A269590.
The digits of the other 5-adic integer sqrt(-4), are given in A269591. See also A268922 for the two 5-adic numbers -u and u.
a(n) is the unique solution of the linear congruence 2*A269590(n)*a(n) + A269594(n) == 0 (mod 5), n>=1. Therefore only the values 0, 1, 2, 3 and 4 appear. See the Nagell reference given in A268922, eq. (6) on p. 86, adapted to this case. a(0) = 4 follows from the formula given below.

			a(4) = -212*(2*364)^3 (mod 5) = -2*(2*(-1))^3 (mod 5) = 1.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
BCMATH Congruence Programs, Finding a p-adic square root of a quadratic residue (mod p), p an odd prime..

Cf. A269590, A269591 (companion), A269594.

a(n) = truncate(-sqrt(-4+O(5^(n+1))))\5^n; \\ Michel Marcus, Mar 04 2016

a(n) = (b(n+1) - b(n))/5^n, n>=0, with b(n) = A269590(n), n >= 0.
a(n) = -A269594(n)*(2*A269590(n))^3 (mod 5), n >= 1. Solution of the linear congruence see, e.g., Nagell, Theorem 38 pp. 77-78.
A269590(n+1) = sum(a(k)*5^k, k=0..n), n>=0.
a(n) = 4 - A269591(n) if n > 0 and a(0) = 5 - A269591(0) = 5-4 = 1. - Michel Marcus, Mar 31 2016

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).

Original entry on oeis.org

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

A271223 Digits of one of the two 3-adic integers sqrt(-2).

Original entry on oeis.org

1, 1, 2, 0, 0, 2, 0, 1, 0, 0, 0, 2, 1, 2, 0, 1, 0, 1, 1, 0, 2, 2, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 2, 1, 0, 1, 2, 0, 2, 2, 0, 2, 0, 1, 2, 0, 1, 2, 2, 2, 1, 0, 2, 0, 1, 2, 0, 2, 0, 0, 1, 1, 2, 1, 0, 1, 2, 1, 1, 2, 0

Offset: 0

Wolfdieter Lang, Apr 05 2016

This is the scaled first difference sequence of A268924. See the formula.
The digits of the other 3-adic integer sqrt(-2), are given in A271224. See also A268924 for the two 3-adic numbers sqrt(-2), called there u and -u.
a(n) is the unique solution of the linear congruence 2*A268924(n)*a(n) + A271225(n) == 0 (mod 3), n>=1. Therefore only the values 0, 1, and 2 appear. See the Nagell reference given in A268922, eq. (6) on p. 86, adapted to this case. a(0) = 1 follows from the formula given below.
For details see the Wolfdieter Lang link under A268992.
The first k digits in the base 3 representation of Lucas(3^n) give the first k terms of the sequence. For example, the base 3 representation of Lucas(3^5) = 84722519070079276 begins 1 + 1*3 + 2*(3^2) + 0*(3^3) + 0*(3^4) + ... so the sequence begins [1, 1, 2, 0, 0, ...]. - Peter Bala, Nov 15 2022

			a(4) = 0 because 2*22*3 + 6 = 138 == 0 (mod 3).
a(4) = - 6*(2*22) (mod 3) = -0*(2*1) (mod 3) = 0.
A268924(5) =  22 = 1*3^0 + 1*3^1 + 2*3^2 + 0*3^3 + 0*3^4.

Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, pp. 86 and 77-78.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
BCMATH Congruence Programs, Finding a p-adic square root of a quadratic residue (mod p), p an odd prime.

Cf. A006267, A268924, A268992, A271224, A271225.

# uses properties of the numbers Lucas(3^n) = A006267(n)
a := proc(n) option remember; if n = 1 then 1 else irem(a(n-1)^3 + 3*a(n-1), 3^n) end if; end proc:
convert(a(70), base, 3); # Peter Bala, Nov 15 2022

a(n) = truncate(sqrt(-2+O(3^(n+1))))\3^n; \\ Michel Marcus, Apr 09 2016

a(n) = (b(n+1) - b(n))/3^n, n >= 0, with b(n) = A268924(n), n >= 0.
a(n) = - A271225(n)*2*A268924(n) (mod 3), n >= 1. Solution of the linear congruence given above in a comment. See, e.g., Nagell, Theorem 38 pp. 77-78.
A268924(n+1) = sum(a(k)*3^k, k=0..n), n >= 0.

A268924 One of the two successive approximations up to 3^n for the 3-adic integer sqrt(-2). These are the numbers congruent to 1 mod 3 (except for n = 0).

Original entry on oeis.org

0, 1, 4, 22, 22, 22, 508, 508, 2695, 2695, 2695, 2695, 356989, 888430, 4077076, 4077076, 18425983, 18425983, 147566146, 534986635, 534986635, 7508555437, 28429261843, 28429261843, 122572440670, 405001977151

Offset: 0

Wolfdieter Lang, Apr 05 2016

The other approximation for the 3-adic integer sqrt(-2) with numbers 2 (mod 3) is given in A271222.
For the digits of this 3-adic integer sqrt(-2), that is the scaled first differences, see A271223. This 3-adic number has the digits read from the right to the left ...2202101200022211102201101021200010200211 = u.
The companion 3-adic number has digits ...20020121022200011120021121201022212022012 = -u. See A271224.
For details see the W. Lang link ``Note on a Recurrence or Approximation Sequences of p-adic Square Roots'' given under A268922, also for the Nagell reference and Hensel lifting. Here p = 3, b = 2, x_1 = 1 and  z_1 = 1.

			n=2: 4^2 + 2 = 18 == 0 (mod 3^2), and 4 is the only solution from {0, 1, ..., 8} which is congruent to 1 modulo 3.
n=3: the only solution of  X^2 + 2 == 0 (mod 3^3) with X from {0, ..., 26} and X == 1 (mod 3) is 22. The number 5 = A271222(3)  also satisfies the first congruence but not the second one: 5  == 2 (mod 3).
n=4: the only solution of X^2 + 2 == 0 (mod 3^4) with X from {0, ..., 80} and X == 1 (mod 3) is also 22. The number 59 = A271222(4) also satisfies the first congruence but not the second one: 59  == 2 (mod 3).

Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, p. 87.

Seiichi Manyama, Table of n, a(n) for n = 0..2096
Peter Bala, A note on A268924 and A271222, Nov 28 2022.
Wikipedia, Hensel's Lemma.

Cf. A000032, A006267, A268922, A271222, A271223, A271224.

with(padic):D1:=op(3,op([evalp(RootOf(x^2+2),3,20)][1])): 0,seq(sum('D1[k]*3^(k-1)','k'=1..n), n=1..20);
# alternative program based on the Lucas numbers L(3^n) = A006267(n)
a := proc(n) option remember; if n = 1 then 1 else irem(a(n-1)^3 + 3*a(n-1), 3^n) end if; end: seq(a(n), n = 1..22); # Peter Bala, Nov 15 2022

a(n) = truncate(sqrt(-2+O(3^(n)))); \\ Michel Marcus, Apr 09 2016

def a268924(n):
    ary=[0]
    a, mod = 1, 3
    for i in range(n):
          b=a%mod
          ary.append(b)
          a=b**2 + b + 2
          mod*=3
    return ary
print(a268924(100)) # Indranil Ghosh, Aug 04 2017, after Ruby

def A268924(n)
  ary = [0]
  a, mod = 1, 3
  n.times{
    b = a % mod
    ary << b
    a = b * b + b + 2
    mod *= 3
  }
  ary
end
p A268924(100) # Seiichi Manyama, Aug 03 2017

a(n)^2 + 2 == 0 (mod 3^n), and a(n) == 1 (mod 3). Representatives of the complete residue system {0, 1, ..., 3^n-1} are taken.
Recurrence for n >= 1: a(n) = modp(a(n-1) +  a(n-1)^2 + 2, 3^n), n >= 2, with a(1) = 1. 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) = 3^n - A271222(n), n >= 1.
a(n) == Lucas(3^n) (mod 3^n). - Peter Bala, Nov 10 2022

A271224 Digits of one of the two 3-adic integers sqrt(-2). Here the sequence with first digit 2.

Original entry on oeis.org

2, 1, 0, 2, 2, 0, 2, 1, 2, 2, 2, 0, 1, 0, 2, 1, 2, 1, 1, 2, 0, 0, 2, 1, 1, 1, 0, 0, 0, 2, 2, 2, 0, 1, 2, 1, 0, 2, 0, 0, 2, 0, 2, 1, 0, 2, 1, 0, 0, 0, 1, 2, 0, 2, 1, 0, 2, 0, 2, 2, 1

Offset: 0

Wolfdieter Lang, Apr 05 2016

This is the scaled first difference sequence of A271222. See the formula.
The digits of the other 3-adic integer sqrt(-2), are given in A271223. See also a comment on A268924 for the two 3-adic numbers sqrt(-2), called there u and -u.
a(n) is the unique solution of the linear congruence 2*A271222(n)*a(n) + A271226(n) == 0 (mod 3), n>=1. Therefore only the values 0, 1, and 2 appear. See the Nagell reference given in A268922, eq. (6) on p. 86, adapted to this case.
  a(0) = 2 follows from the formula given below.
For details see the Wolfdieter Lang link under A268992.
The first k digits in the base 3 representation of A002203(3^k) = A006266(k) give the first k terms of the sequence. - Peter Bala, Nov 26 2022

			a(4) = 2 because 2*59*2 + 43 = 279 == 0 (mod 3).
a(4) = - 43*(2*59) (mod 3) = -1*(2*(-1)) (mod 3) = 2.
A271222(5) = 221  = 2*3^0 + 1*3^1 + 0*3^2 + 2*3^3 + 2*3^4.

Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, pp. 86 and 77-78.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Peter Bala, A note on A268924 and A271222, Nov 28 2022.
BCMATH Congruence Programs, Finding a p-adic square root of a quadratic residue (mod p), p an odd prime.

Cf. A268924, A268992, A271222, A271226, A271223 (companion).

a(n) = truncate(-sqrt(-2+O(3^(n+1))))\3^n; \\ Michel Marcus, Apr 09 2016

a(n) = (b(n+1) - b(n))/3^n, n >= 0, with b(n) = A271222(n), n >= 0.
a(n) = - A271226(n)*2*A271222(n) (mod 3), n >= 1. Solution of the linear congruence given above in a comment. See, e.g., Nagell, Theorem 38 pp. 77-78.
A271222(n+1) = sum(a(k)*3^k, k=0..n), n >= 0.

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.

A271222 One of the two successive approximations up to 3^n for the 3-adic integer sqrt(-2). These are the numbers congruent to 2 mod 3 (except for the initial 0).

Original entry on oeis.org

0, 2, 5, 5, 59, 221, 221, 1679, 3866, 16988, 56354, 174452, 174452, 705893, 705893, 10271831, 24620738, 110714180, 239854343, 627274832, 2951797766, 2951797766, 2951797766, 65713916984, 159857095811, 442286632292

Offset: 0

Wolfdieter Lang, Apr 05 2016

The other approximation for the 3-adic integer sqrt(-2) with numbers 1 (mod 3) is given in A268924.
For the digits of this 3-adic integer sqrt(-2), that is the scaled first differences, see A271224. This 3-adic number has the digits read from the right to the left ... 20020121022200011120021121201022212022012 = -u. For the digits of u see A271223.
For details see the W. Lang link ``Note on a Recurrence or Approximation Sequences of p-adic Square Roots'' given under A268922, also for the Nagell reference and Hensel lifting. Here p = 3,  b = 2, x_2 = 2 and  z_2 = 2.

			n=2: 5^2 + 2 = 27 == 0 (mod 3^2), and 5 is the only solution from {0, 1, ..., 8} which is congruent to 2 modulo 3.
n=3: the only solution of X^2 + 2 == 0 (mod 3^3) with X from {0, ..., 26} and X == 2(mod 3) is 5. The number 22 = A268924(3) also satisfies the first congruence but not the second one: 22  == 1 (mod 3).
n=4: the only solution of X^2 + 2 == 0 (mod 3^4) with X from {0, ..., 80} and X == 2 (mod 3) is 59. The number 22 = A268924(4) also satisfies the first congruence but not the second one: 59  == 1 (mod 3).

Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, p. 87.

Seiichi Manyama, Table of n, a(n) for n = 0..2095
Peter Bala, A note on A268924 and A271222, Nov 28 2022.
Wikipedia, Hensel's Lemma.

Cf. A002293, A268922, A268924, A271223, A271224.

with(padic):D2:=op(3,op([evalp(RootOf(x^2+2),3,20)][2])): 0,seq(sum('D2[k]*3^(k-1)','k'=1..n), n=1..20);

a(n) = if (n, 3^n - truncate(sqrt(-2+O(3^(n)))), 0); \\ Michel Marcus, Apr 09 2016

def a271222(n):
      ary=[0]
      a, mod = 2, 3
      for i in range(n):
          b=a%mod
          ary.append(b)
          a=2*b**2 + b + 4
          mod*=3
      return ary
print(a271222(100)) # Indranil Ghosh, Aug 04 2017, after Ruby

def A271222(n)
  ary = [0]
  a, mod = 2, 3
  n.times{
    b = a % mod
    ary << b
    a = 2 * b * b + b + 4
    mod *= 3
  }
  ary
end
p A271222(100) # Seiichi Manyama, Aug 03 2017

a(n)^2 + 2 == 0 (mod 3^n), and a(n) == 2 (mod 3). Representatives of the complete residue system {0, 1, ..., 3^n-1} are taken.
Recurrence for n >= 1: a(n) = modp(a(n-1) +  2*(a(n-1)^2 + 2), 3^n), n >= 2, with a(1) = 2. 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) = 3^n - A268924(n), n >= 1.
a(n) == A002203(3^n) (mod 3^n). - Peter Bala, Nov 10 2022

cp's OEIS Frontend

A269593 a(n) = (A268922(n)^2 + 4)/5^n, n >= 0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

Extensions

A048898 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-1).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

Extensions

A269591 Digits of one of the two 5-adic integers sqrt(-4).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A269592 Digits of one of the two 5-adic integers sqrt(-4). Here the ones related to A269590.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

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).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

PARI

Formula

A271223 Digits of one of the two 3-adic integers sqrt(-2).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

PARI

Formula

A268924 One of the two successive approximations up to 3^n for the 3-adic integer sqrt(-2). These are the numbers congruent to 1 mod 3 (except for n = 0).

Views

Author

Keywords

Comments