A269590 -id:A269590 - cp's OEIS frontend

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

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

A269594 a(n) = (A269590(n)^2 + 4)/5^n, n >= 0.

Original entry on oeis.org

4, 4, 8, 104, 212, 313, 11645, 2329, 73757, 160772, 646925, 129385, 25877, 49838696, 2503218301, 16177487972, 18226737365, 3645347473, 2526514341077, 2510040201736, 43137313790909, 136233128831473, 1960924754787877, 1733911367978596, 27260145118408781

Offset: 0

Wolfdieter Lang, Mar 02 2016

a(n) is an integer because b(n) = A269590(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)= (364^2 + 4)/5^4 = 212.

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

Cf. A268922, A269590, A269593 (companion).

b(n) = if (n==0, 0, 5^n - truncate(sqrt(-4+O(5^(n)))));
a(n) = (b(n)^2 + 4)/5^n; \\ Michel Marcus, Mar 24 2016

a(n) = (b(n)^2 + 4)/5^n, n>=0, with b(n) = A269590(n).

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

A268922 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 1 mod 5 numbers, except for n = 0.

Original entry on oeis.org

0, 1, 11, 11, 261, 2136, 2136, 64636, 220886, 1392761, 7252136, 46314636, 241627136, 974049011, 2194752136, 8298267761, 99851002136, 710202564636, 710202564636, 12154294361511, 31227780689636

Offset: 0

Wolfdieter Lang, Mar 02 2016

The other approximation for the 5-adic integer sqrt(-4) with numbers 4 (mod 5) is given in A269590.
For the two approximations of the 5-adic integer sqrt(-1) see A048899 and A048898. For comments and some proofs see A210848.
For the digits of this 5-adic integer sqrt(-4), that is the scaled first differences, see A269591. This 5-adic number has the digits read from the right to the left ...32234111132111101130213043113443324032021 = u.
  The companion 5-adic number -u has digits ....12210333312333343314231401331001120412424. See A269592.
The recurrence given below (for n >= 1) has been derived from the following facts (i) x^2 + 4 == 0 (mod 5) has the two distinct solutions x(1) = 1 and x(2) = 5 - x(1) = 4. This guarantees the existence of a unique solution x = x1(n) of  x^2 + 4 == 0 (mod 5^n) , for n >= 2 , which satisfies also x1(n) == 1 (mod 5). See e.g., Theorem 50, p. 87 of the Nagell reference. The same is true for the solution x = x2(n) with x2(n) == 4 (mod 5^n). (ii) As a consequence of Hensel's lemma (see e.g., the Wikipedia reference under Hensel lifting) one knows that  x1(n) (which is treated here) satisfies the congruence x1(n) == x1(n-1) (mod 5^(n-1)) with x1(1) = x1 = 1. (A similar statement holds for x2(n) with input x2(1) = x(2) = 4. This is used in A269590). These two facts allow one to derive a recurrence for x1(n) (and for x2(n)).

			n=2:  11^2 + 4 = 125  == 0 (mod 5^2), and 125 is the only solution from {0, 1, ..., 24} which is congruent to 1 modulo 5.
n=3:  the only solution of x1^2 + 4 == 0 (mod 5^3) with x1 from {0, ..., 124} and x1 ==  1 (mod 4) is also 11. The number 114 satisfies also the first congruence but not the second one: 114 == 2 (mod 4).

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

Seiichi Manyama, Table of n, a(n) for n = 0..1430
Wolfdieter Lang, Note on a Recurrence for Approximation Sequences of p-adic Square Roots
Wikipedia, Hensel's lemma.

Cf. A000032, A048899, A048898, A144837, A210848, A269590, A269591, A269592.

with(padic):D1:=op(3,op([evalp(RootOf(x^2+4),5,20)][1])): 0,seq(sum('D1[k]*5^(k-1)','k'=1..n), n=1..20);
# alternative program - see A144837
a := proc (n) option remember; if n = 1 then 1 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) = truncate(sqrt(-4+O(5^(n)))); \\ Michel Marcus, Mar 04 2016

Recurrence for n >= 1: a(n) = modp( a(n-1) + 2*(a(n-1)^2 + 4), 5^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) = 5^n - A269590(n), n >= 1.
a(n) == Lucas(5^n) (mod 5^n). - Peter Bala, Nov 10 2022

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.

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.

A318960 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 1 (mod 4) case.

Original entry on oeis.org

1, 5, 5, 21, 53, 53, 181, 181, 181, 181, 181, 181, 181, 16565, 49333, 49333, 49333, 49333, 573621, 1622197, 1622197, 1622197, 10010805, 10010805, 10010805, 77119669, 211337397, 479772853, 479772853, 479772853, 2627256501, 6922223797, 15512158389, 15512158389

Offset: 2

Jianing Song, Sep 06 2018

nonn

a(n) is the unique number k in [1, 2^n] and congruent to 1 (mod 4) such that k^2 + 7 is divisible by 2^(n+1).

The 2-adic integers are very different from p-adic ones where p is an odd prime. For example, provided that there is at least one solution, the number of solutions to x^n = a over p-adic integers is gcd(n, p-1) for odd primes p and gcd(n, 2) for p = 2. For odd primes p, x^2 = a is solvable iff a is a quadratic residue modulo p, while for p = 2 it's solvable iff a == 1 (mod 8). If gcd(n, p-1) > 1 and gcd(a, p) = 1, then the solutions to x^n = a differ starting at the rightmost digit for odd primes p, while for p = 2 they differ starting at the next-to-rightmost digit. As a result, the formulas and the program here are different from those in other entries related to p-adic integers.

			The unique number k in [1, 4] and congruent to 1 modulo 4 such that k^2 + 7 is divisible by 8 is 1, so a(2) = 1.
a(2)^2 + 7 = 8 which is not divisible by 16, so a(3) = a(2) + 2^2 = 5.
a(3)^2 + 7 = 32 which is divisible by 32, so a(4) = a(3) = 5.
a(4)^2 + 7 = 32 which is divisible by 64, so a(5) = a(4) + 2^4 = 21.
a(5)^2 + 7 = 448 which is divisible by 128, so a(6) = a(5) + 2^5 = 53.
...

Jianing Song, Table of n, a(n) for n = 2..999 (offset corrected by Jianing Song)
G. P. Michon, Introduction to p-adic integers, Numericana.

Cf. A318962.
Expansions of p-adic integers:
this sequence, A318961 (2-adic, sqrt(-7));
A268924, A271222 (3-adic, sqrt(-2));
A268922, A269590 (5-adic, sqrt(-4));
A048898, A048899 (5-adic, sqrt(-1));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A290800, A290802 (7-adic, sqrt(-6));
A290806, A290809 (7-adic, sqrt(-5));
A290803, A290804 (7-adic, sqrt(-3));
A210852, A212153 (7-adic, (1+sqrt(-3))/2);
A290557, A290559 (7-adic, sqrt(2));
A286840, A286841 (13-adic, sqrt(-1));
A286877, A286878 (17-adic, sqrt(-1)).
Also expansions of 10-adic integers:
A007185, A010690 (nontrivial roots to x^2-x);
A216092, A216093, A224473, A224474 (nontrivial roots to x^3-x).

PARI
```
a(n) = truncate(-sqrt(-7+O(2^(n+1))))
```

a(2) = 1; for n >= 3, a(n) = a(n-1) if a(n-1)^2 + 7 is divisible by 2^(n+1), otherwise a(n-1) + 2^(n-1).
a(n) = 2^n - A318961(n).
a(n) = Sum_{i=0..n-1} A318962(i)*2^i.

Offset corrected by Jianing Song, Aug 28 2019

A318961 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 3 (mod 4) case.

Original entry on oeis.org

3, 3, 11, 11, 11, 75, 75, 331, 843, 1867, 3915, 8011, 16203, 16203, 16203, 81739, 212811, 474955, 474955, 474955, 2572107, 6766411, 6766411, 23543627, 57098059, 57098059, 57098059, 57098059, 593968971, 1667710795, 1667710795, 1667710795, 1667710795, 18847579979

Offset: 2

Jianing Song, Sep 06 2018

nonn

a(n) is the unique number k in [1, 2^n] and congruent to 3 (mod 4) such that k^2 + 7 is divisible by 2^(n+1).

The 2-adic integers are very different from p-adic ones where p is an odd prime. For example, provided that there is at least one solution, the number of solutions to x^n = a over p-adic integers is gcd(n, p-1) for odd primes p and gcd(n, 2) for p = 2. For odd primes p, x^2 = a is solvable iff a is a quadratic residue modulo p, while for p = 2 it's solvable iff a == 1 (mod 8). If gcd(n, p-1) > 1 and gcd(a, p) = 1, then the solutions to x^n = a differ starting at the rightmost digit for odd primes p, while for p = 2 they differ starting at the next-to-rightmost digit. As a result, the formulas and the program here are different from those in other entries related to p-adic integers.

			The unique number k in [1, 4] and congruent to 3 modulo 4 such that k^2 + 7 is divisible by 8 is 3, so a(2) = 3.
a(2)^2 + 7 = 16 which is divisible by 16, so a(3) = a(2) = 3.
a(3)^2 + 7 = 16 which is not divisible by 32, so a(4) = a(3) + 2^3 = 11.
a(4)^2 + 7 = 128 which is divisible by 64, so a(5) = a(4) = 11.
a(5)^2 + 7 = 128 which is divisible by 128, so a(6) = a(5) = 11.
...

Jianing Song, Table of n, a(n) for n = 2..999 (offset corrected by Jianing Song)
G. P. Michon, Introduction to p-adic integers, Numericana.

Cf. A318963.
Expansions of p-adic integers:
A318960, this sequence (2-adic, sqrt(-7));
A268924, A271222 (3-adic, sqrt(-2));
A268922, A269590 (5-adic, sqrt(-4));
A048898, A048899 (5-adic, sqrt(-1));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A290800, A290802 (7-adic, sqrt(-6));
A290806, A290809 (7-adic, sqrt(-5));
A290803, A290804 (7-adic, sqrt(-3));
A210852, A212153 (7-adic, (1+sqrt(-3))/2);
A290557, A290559 (7-adic, sqrt(2));
A286840, A286841 (13-adic, sqrt(-1));
A286877, A286878 (17-adic, sqrt(-1)).
Also expansions of 10-adic integers:
A007185, A010690 (nontrivial roots to x^2-x);
A216092, A216093, A224473, A224474 (nontrivial roots to x^3-x).

a(n) = if(n==2, 3, truncate(sqrt(-7+O(2^(n+1)))))

a(2) = 3; for n >= 3, a(n) = a(n-1) if a(n-1)^2 + 7 is divisible by 2^(n+1), otherwise a(n-1) + 2^(n-1).
a(n) = 2^n - A318960(n).
a(n) = Sum_{i=0..n-1} A318963(i)*2^i.

Offset corrected by Jianing Song, Aug 28 2019

A324027 One of the two successive approximations up to 5^n for 5-adic integer sqrt(-6). This is the 2 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 2, 12, 37, 162, 1412, 10787, 42037, 354537, 1526412, 3479537, 3479537, 3479537, 247620162, 3909729537, 10013245162, 101565979537, 711917542037, 2237796448287, 13681888245162, 51828860901412, 337931155823287, 1291605472229537, 10828348636292037, 58512064456604537

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 2 modulo 5.

A324028 is the approximation (congruent to 3 mod 5) of another square root of 6 over the 5-adic field.

			12^2 = 144 = 6*5^2 - 6;
37^2 = 1369 = 11*5^3 - 6;
162^2 = 26244 = 42*5^4 - 6.

Wikipedia, p-adic number

Cf. A048898, A048899, A324029, A324030.
Approximations of 5-adic square roots:
this sequence, A324028 (sqrt(-6));
A268922, A269590 (sqrt(-4));
A048898, A048899 (sqrt(-1));
A324023, A324024 (sqrt(6)).

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

For n > 0, a(n) = 5^n - A324028(n).

a(n) = A048898(n)*A324023(n) mod 5^n = A048899(n)*A324024(n) mod 5^n.

A324028 One of the two successive approximations up to 5^n for 5-adic integer sqrt(-6). This is the 3 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 3, 13, 88, 463, 1713, 4838, 36088, 36088, 426713, 6286088, 45348588, 240661088, 973082963, 2193786088, 20504332963, 51021911088, 51021911088, 1576900817338, 5391598082963, 43538570739213, 138906002379838, 1092580318786088, 1092580318786088, 1092580318786088

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 3 modulo 5.

A324027 is the approximation (congruent to 3 mod 5) of another square root of -6 over the 5-adic field.

			13^2 = 169 = 7*5^2 - 6;
88^2 = 7744 = 62*5^3 - 6;
463^2 = 214369 = 343*5^4 - 6.

Wikipedia, p-adic number

Cf. A048898, A048899, A324029, A324030.
Approximations of 5-adic square roots:
A324027, sequence (sqrt(-6));
A268922, A269590 (sqrt(-4));
A048898, A048899 (sqrt(-1));
A324023, A324024 (sqrt(6)).

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

For n > 0, a(n) = 5^n - A324027(n).

a(n) = A048898(n)*A324024(n) mod 5^n = A048899(n)*A324023(n) mod 5^n.

cp's OEIS Frontend

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

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A269594 a(n) = (A269590(n)^2 + 4)/5^n, n >= 0.

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A268922 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 1 mod 5 numbers, except for n = 0.

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Maple

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

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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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A318960 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 1 (mod 4) case.

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