cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A324025 Digits of one of the two 5-adic integers sqrt(6) that is related to A324023.

Original entry on oeis.org

1, 3, 0, 4, 2, 1, 2, 3, 1, 3, 3, 0, 3, 3, 2, 3, 2, 2, 2, 4, 3, 3, 1, 4, 0, 1, 2, 0, 0, 0, 3, 3, 1, 4, 1, 0, 1, 2, 4, 1, 4, 1, 1, 0, 2, 4, 4, 3, 0, 2, 3, 4, 1, 1, 4, 3, 4, 2, 4, 2, 1, 1, 2, 4, 4, 3, 2, 3, 1, 1, 0, 1, 4, 2, 3, 4, 4, 4, 4, 0, 3, 3, 1, 2, 3, 2, 3, 1
Offset: 0

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

This square root of 6 in the 5-adic field ends with digit 1. The other, A324026, ends with digit 4.

Examples

			The solution to x^2 == 6 (mod 5^4) such that x == 1 (mod 5) is x == 516 (mod 5^4), and 516 is written as 4031 in quinary, so the first four terms are 1, 3, 0 and 4.

Links

Crossrefs

Cf. A324023, A324024.
Digits of 5-adic square roots:
A324029, A324030 (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
this sequence, A324026 (sqrt(6)).

Programs

  • PARI
    a(n) = truncate(sqrt(6+O(5^(n+1))))\5^n

Formula

a(n) = (A324023(n+1) - A324023(n))/5^n.
For n > 0, a(n) = 4 - A324026(n).
Equals A210850*A324030 = A210851*A324029, where each A-number represents a 5-adic number.

A324026 Digits of one of the two 5-adic integers sqrt(6) that is related to A324024.

Original entry on oeis.org

4, 1, 4, 0, 2, 3, 2, 1, 3, 1, 1, 4, 1, 1, 2, 1, 2, 2, 2, 0, 1, 1, 3, 0, 4, 3, 2, 4, 4, 4, 1, 1, 3, 0, 3, 4, 3, 2, 0, 3, 0, 3, 3, 4, 2, 0, 0, 1, 4, 2, 1, 0, 3, 3, 0, 1, 0, 2, 0, 2, 3, 3, 2, 0, 0, 1, 2, 1, 3, 3, 4, 3, 0, 2, 1, 0, 0, 0, 0, 4, 1, 1, 3, 2, 1, 2, 1, 3
Offset: 0

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

This square root of 6 in the 5-adic field ends with digit 4. The other, A324025, ends with digit 1.

Examples

			The solution to x^2 == 6 (mod 5^4) such that x == 4 (mod 5) is x == 109 (mod 5^4), and 109 is written as 414 in quinary, so the first four terms are 4, 1, 4 and 0.

Links

Crossrefs

Cf. A324023, A324024.
Digits of 5-adic square roots:
A324029, A324030 (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
A324025, this sequence (sqrt(6)).

Programs

  • PARI
    a(n) = truncate(-sqrt(6+O(5^(n+1))))\5^n

Formula

a(n) = (A324024(n+1) - A324024(n))/5^n.
For n > 0, a(n) = 4 - A324025(n).
Equals A210850*A324029 = A210851*A324030, where each A-number represents a 5-adic number.

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

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

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.

Examples

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

Links

Crossrefs

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)).

Programs

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

Formula

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

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

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.

Examples

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

Links

Crossrefs

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)).

Programs

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

Formula

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.

A324029 Digits of one of the two 5-adic integers sqrt(-6) that is related to A324027.

Original entry on oeis.org

2, 2, 1, 1, 2, 3, 2, 4, 3, 1, 0, 0, 1, 3, 1, 3, 4, 2, 3, 2, 3, 2, 4, 4, 2, 3, 3, 0, 1, 1, 3, 1, 1, 1, 3, 1, 2, 3, 2, 3, 4, 1, 0, 2, 4, 4, 3, 4, 0, 3, 2, 0, 2, 0, 2, 0, 3, 2, 0, 0, 4, 2, 4, 4, 0, 4, 4, 4, 3, 1, 4, 2, 2, 4, 2, 0, 0, 0, 3, 0, 4, 3, 2, 4, 3, 3, 4, 0
Offset: 0

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

This square root of -6 in the 5-adic field ends with digit 2. The other, A324030, ends with digit 3.

Examples

			The solution to x^2 == -6 (mod 5^4) such that x == 2 (mod 5) is x == 162 (mod 5^4), and 162 is written as 1122 in quinary, so the first four terms are 2, 2, 1 and 1.

Links

Crossrefs

Cf. A324027, A324028.
Digits of 5-adic square roots:
this sequence, A324030 (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
A324025, A324026 (sqrt(6)).

Programs

  • PARI
    a(n) = truncate(sqrt(-6+O(5^(n+1))))\5^n

Formula

a(n) = (A324027(n+1) - A324027(n))/5^n.
For n > 0, a(n) = 4 - A324030(n).
Equals A210850*A324026 = A210851*A324025, where each A-number represents a 5-adic number.

A327305 Digits of one of the two 5-adic integers sqrt(-9) that is related to A327303.

Original entry on oeis.org

4, 0, 3, 0, 0, 1, 1, 4, 2, 0, 2, 2, 3, 2, 4, 4, 1, 1, 2, 2, 3, 0, 2, 2, 4, 2, 1, 4, 1, 4, 0, 0, 0, 2, 4, 1, 1, 3, 1, 1, 0, 4, 1, 2, 1, 2, 2, 1, 1, 2, 0, 0, 3, 1, 2, 0, 4, 2, 0, 3, 4, 4, 0, 0, 0, 0, 1, 4, 0, 3, 4, 0, 1, 4, 4, 3, 3, 0, 2, 3, 2, 3, 3, 3, 1, 4, 2, 4
Offset: 0

Views

Author

Jianing Song, Sep 16 2019

Keywords

Comments

This is the 5-adic solution to x^2 = -9 that ends in 4. A327304 gives the other solution that ends in 1.

Examples

			Equals ...1131142000414124220322114423220241100304.

Links

Crossrefs

Cf. A327302, A327303.
Digits of 5-adic square roots:
A327304, this sequence (sqrt(-9));
A324029, A324030 (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
A324025, A324026 (sqrt(6)).

Programs

  • Maple
    op([1,1,3], select(t -> padic:-ratvaluep(t,1)=4, [padic:-rootp(x^2+9,5,100)])); # Robert Israel, Aug 31 2020
  • PARI
    a(n) = truncate(-sqrt(-9+O(5^(n+1))))\5^n

Formula

For n > 0, a(n) is the unique m in {0, 1, 2, 3, 4} such that (A327303(n) + m*5^n)^2 + 9 is divisible by 5^(n+1).
a(n) = (A327303(n+1) - A327303(n))/5^n.
For n > 0, a(n) = 4 - A327304(n).

A327304 Digits of one of the two 5-adic integers sqrt(-9) that is related to A327302.

Original entry on oeis.org

1, 4, 1, 4, 4, 3, 3, 0, 2, 4, 2, 2, 1, 2, 0, 0, 3, 3, 2, 2, 1, 4, 2, 2, 0, 2, 3, 0, 3, 0, 4, 4, 4, 2, 0, 3, 3, 1, 3, 3, 4, 0, 3, 2, 3, 2, 2, 3, 3, 2, 4, 4, 1, 3, 2, 4, 0, 2, 4, 1, 0, 0, 4, 4, 4, 4, 3, 0, 4, 1, 0, 4, 3, 0, 0, 1, 1, 4, 2, 1, 2, 1, 1, 1, 3, 0, 2, 0
Offset: 0

Views

Author

Jianing Song, Sep 16 2019

Keywords

Comments

This is the 5-adic solution to x^2 = -9 that ends in 1. A327305 gives the other solution that ends in 4.

Examples

			Equals ...3313302444030320224122330021224203344141.

Links

Crossrefs

Cf. A327302, A327303.
Digits of 5-adic square roots:
this sequence, A327305 (sqrt(-9));
A324029, A324030 (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
A324025, A324026 (sqrt(6)).

Programs

  • PARI
    a(n) = truncate(-sqrt(-9+O(5^(n+1))))\5^n

Formula

For n > 0, a(n) is the unique m in {0, 1, 2, 3, 4} such that (A327302(n) + m*5^n)^2 + 9 is divisible by 5^(n+1).
a(n) = (A327302(n+1) - A327302(n))/5^n.
For n > 0, a(n) = 4 - A327305(n).
Showing 1-7 of 7 results.

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