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 1-9 of 9 results.

A325490 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 2 mod 5.

Original entry on oeis.org

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

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

One of the two square roots of A324026, where an A-number represents a 5-adic number. The other square root is A325491.
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.

Examples

			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 k = 22 = (42)_5, so the first three terms are 2, 4 and 0.

Links

Crossrefs

Cf. A210850, A210851, A324026, A325484, A325485, A325486, A325487.
Digits of p-adic fourth-power roots:
A325489, this sequence, A325491, A325492 (5-adic, 6^(1/4));
A324085, A324086, A324087, A324153 (13-adic, 3^(1/4)).

Programs

  • Maple
    S:= select(t -> op([1,3,1],t)=2, [padic:-rootp(_Z^4-6,5,100)]):
op([1,1,3],S); # Robert Israel, Mar 23 2023
  • PARI
    a(n) = lift(sqrtn(6+O(5^(n+1)), 4) * sqrt(-1+O(5^(n+1))))\5^n

Formula

Equals A325489*A210850 = A325492*A210851.
a(n) = (A325485(n+1) - A325485(n))/13^n.
For n > 0, a(n) = 4 - A325491(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

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 1 modulo 5.
A324024 is the approximation (congruent to 4 mod 5) of another square root of 6 over the 5-adic field.

Examples

			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.

Links

Crossrefs

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)).

Programs

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

Formula

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

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 1 modulo 5.
A324023 is the approximation (congruent to 4 mod 5) of another square root of 6 over the 5-adic field.

Examples

			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.

Links

Crossrefs

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)).

Programs

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

Formula

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.

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.

A325491 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 3 mod 5.

Original entry on oeis.org

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

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

One of the two square roots of A324026, where an A-number represents a 5-adic number. The other square root is A325490.
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.

Examples

			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 = (403)_5, so the first three terms are 3, 0 and 4.

Links

Crossrefs

Cf. A210850, A210851, A324026, A325484, A325485, A325486, A325487.
Digits of p-adic fourth-power roots:
A325489, A325490, this sequence, A325492 (5-adic, 6^(1/4));
A324085, A324086, A324087, A324153 (13-adic, 3^(1/4)).

Programs

  • PARI
    a(n) = lift(-sqrtn(6+O(5^(n+1)), 4) * sqrt(-1+O(5^(n+1))))\5^n

Formula

Equals A325489*A210851 = A325492*A210850.
a(n) = (A325486(n+1) - A325486(n))/13^n.
For n > 0, a(n) = 4 - A325490(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.

A324030 Digits of one of the two 5-adic integers sqrt(-6) that is related to A324028.

Original entry on oeis.org

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

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

This square root of -6 in the 5-adic field ends with digit 3. The other, A324029, ends with digit 2.

Examples

			The solution to x^2 == -6 (mod 5^4) such that x == 3 (mod 5) is x == 463 (mod 5^4), and 463 is written as 3323 in quinary, so the first four terms are 3, 2, 3 and 3.

Links

Crossrefs

Cf. A324027, A324028.
Digits of 5-adic square roots:
A324029, sequence (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) = (A324028(n+1) - A324028(n))/5^n.
For n > 0, a(n) = 4 - A324029(n).
Equals A210850*A324025 = A210851*A324026, 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 1-9 of 9 results.

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