A268922 -id:A268922 - cp's OEIS frontend

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).

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.

A144837 a(n) = Lucas(5^n).

Original entry on oeis.org

11, 167761, 132878596168524201724674011

Offset: 1

Artur Jasinski, Sep 22 2008

Previous name was: a(n) = round(phi^(5^n)) where phi = 1.6180339887... = (sqrt(5) + 1)/2 = A001622.

a(4), a 131-digit number, is too large to show here.

			The base 5 representation of a(3) = 132878596168524201724674011 begins 1 + 2*5 + 0*(5^2) + 2*(5^3) + 3*(5^4) + 0*(5^5) + 4*(5^6) + O(5^7) so A269591 begins [1, 2, 0, 2, 3, 0, 4, ...]. - _Peter Bala_, Nov 14 2022

Amiram Eldar, Table of n, a(n) for n = 1..5

Cf. A000032, A000351, A001622, A001566, A006267, A144836, A144838, A144839, A268922, A269591.

a := proc(n) option remember; if n = 1 then 11 else a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1) end if; end;
seq(a(n), n = 1..5); # Peter Bala, Nov 14 2022

Table[Round[GoldenRatio^(5^n)], {n, 1, 5}]
c = (1 + Sqrt[5])/2; Table[Expand[c^(5^n) + (1 - c)^(5^n)], {n, 1, 5}] (* Artur Jasinski, Oct 05 2008 *)
LucasL[5^Range[5]] (* Harvey P. Dale, Apr 01 2023 *)

a(n) = phi^(5^n) + (1-phi)^(5^n) = phi^(5^n) + (-phi)^(-5^n). - Artur Jasinski, Oct 05 2008
From Peter Bala, Nov 14 2022: (Start)
a(n) = A000032(5^n).
a(n) = a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1) with a(1) = 11.
a(n) == 1 (mod 5).
a(n+1) == a(n) (mod 5^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).
Conjecture: a(n+1) == a(n) (mod 5^(n+r+1)) for n >= r.
The smallest positive residue of a(n) mod(5^n) = A268922(n).
In the ring of 5-adic integers the limit_{n -> oo} a(n) exists and is equal to A269591. An example is given below. (End)

New name from Peter Bala, Nov 10 2022

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.

A327303 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-9). This is the 4 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 4, 4, 79, 79, 79, 3204, 18829, 331329, 1112579, 1112579, 20643829, 118300079, 850721954, 3292128204, 27706190704, 149776503204, 302364393829, 1065303846954, 8694698378204, 46841671034454, 332943965956329, 332943965956329, 5101315547987579, 28943173458143829

Offset: 0

Jianing Song, Sep 16 2019

nonn

a(n) is the unique number k in [1, 5^n] and congruent to 4 mod 5 such that k^2 + 9 is divisible by 5^n.

			The unique number k in {4, 9, 14, 19, 24} such that k^2 + 9 is divisible by 25 is k = 4, so a(2) = 4.
The unique number k in {4, 29, 54, 79, 104} such that k^2 + 9 is divisible by 125 is k = 79, so a(3) = 46.
The unique number k in {79, 204, 329, 454, 579} such that k^2 + 9 is divisible by 625 is k = 79, so a(4) = 79.

Robert Israel, Table of n, a(n) for n = 0..1424
G. P. Michon, Introduction to p-adic integers, Numericana.

For the digits of sqrt(-9) see A327304 and A327305.
Approximations of 5-adic square roots:
A327302, this sequence (sqrt(-9));
A324027, A324028 (sqrt(-6));
A268922, A269590 (sqrt(-4));
A048898, A048899 (sqrt(-1));
A324023, A324024 (sqrt(6)).

R:= [padic:-rootp(x^2+9,5,101)]:
R:= op(select(t -> padic:-ratvaluep(t,1)=4, R)):
seq(padic:-ratvaluep(R,n),n=0..100); # Robert Israel, Jan 16 2023

PARI
```
a(n) = truncate(-sqrt(-9+O(5^n)))
```

a(1) = 4; for n >= 2, a(n) is the unique number k in {a(n-1) + m*5^(n-1) : m = 0, 1, 2, 3, 4} such that k^2 + 9 is divisible by 5^n.

For n > 0, a(n) = 5^n - A327302(n).

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

A269596 Irregular triangle giving in row n the smaller of the two roots x1 of x^2 + b modulo prime(n) from {0, 1, ..., prime(n)-1} corresponding to b from row n of A269595.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 3, 4, 2, 5, 1, 5, 6, 3, 2, 4, 1, 4, 7, 8, 3, 5, 2, 6, 1, 6, 4, 7, 3, 8, 5, 9, 2, 1, 8, 4, 6, 9, 3, 10, 11, 2, 7, 5, 1, 12, 5, 13, 9, 14, 7, 4, 10, 3, 6, 8, 11, 2, 1

Offset: 1

Wolfdieter Lang, Apr 03 2016

The length of row n >= 2 is (prime(n)-1)/2 =  A005097(n-1), and for row n = 1 it is 1.
The other roots of x^2 + b modulo prime(n) are given in A269597.
See A269595 for the irregular triangle with the quadratic residues -b modulo prime(n) = A000040(n), for n >= 1.  For n=1 (prime 2) there is a double root x1 = x2 = 1 of x^2 + 1 (mod 2).
Each row n >= 2 consists of a certain permutation of  1, 2, ..., (prime(n)-1)/2.
For a(n), n >= 2, see column x_1 of the table in the Wolfdieter Lang link.

			The irregular triangle T(n, k) begins (P(n) stands here for prime(n)):
n, P(n)\k 1  2  3  4  5  6  7  8  9 10 11 12 13 14
1,   2:   1
2,   3:   1
3,   5:   2  1
4,   7:   2  3  1
5,  11:   3  4  2  5  1
6:  13:   5  6  3  2  4  1
7,  17:   4  7  8  3  5  2  6  1
8,  19:   6  4  7  3  8  5  9  2  1
9,  23:   8  4  6  9  3 10 11  2  7  5  1
10, 29:  12  5 13  9 14  7  4 10  3  6  8 11  2  1
...
Row n=7 (prime 17) is the permutation (in cycle notation) (1,4,3,8)(2,7,6) of {1, 2, ..., 8}.

Wolfdieter Lang, Note on a Recurrence for Approximation Sequences of p-adic Square Roots

Cf. A000040, A005097, A269595, A269597 (x2).

nn = 12; s = Table[Select[Range[Prime@ n - 1], JacobiSymbol[#, Prime@ n] == 1 &], {n, nn}]; t = Table[Prime@ n - s[[n, (Prime@ n - 1)/2 - k + 1]], {n, Length@ s}, {k, (Prime@ n - 1)/2}] /. {} -> {1};
Prepend[Table[SelectFirst[Range@ #, Function[x, Mod[x^2 + t[[n, k]], #] == 0]] &@ Prime@ n, {n, 2, Length@ t}, {k, (Prime@ n - 1)/2}], {1}] // Flatten (* Michael De Vlieger, Apr 04 2016, Version 10 *)

T(n, k) gives the smaller zero of x^2 + A269595(n, k) == 0 (mod prime(n)), n >= 1, for k=1 if n=1 and k = 1, 2, ..., (prime(n)-1)/2 = A005097(n-1) for n >= 2. Representatives are taken from the complete residue class {0, 1 ,..., prime(n)-1}.

T(n, k) = prime(n) - A269597(n, k).

cp's OEIS Frontend

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A144837 a(n) = Lucas(5^n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

Extensions

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs