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.

A318962 Digits of one of the two 2-adic integers sqrt(-7) that ends in 01.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1
Offset: 0

Views

Author

Jianing Song, Sep 06 2018

Keywords

Comments

Over the 2-adic integers there are 2 solutions to x^2 = -7, one ends in 01 and the other ends in 11. This sequence gives the former one. See A318960 for detailed information.

Examples

			...10110001110011100100110001100000010110101.

Links

Crossrefs

Cf. A318960.
Digits of p-adic integers:
this sequence, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A212152, A212155 (7-adic, (1+sqrt(-3))/2);
A051277, A290558 (7-adic, sqrt(2));
A286838, A286839 (13-adic, sqrt(-1));
A309989, A309990 (17-adic, sqrt(-1)).
Also there are numerous sequences related to digits of 10-adic integers.

Programs

  • PARI
    a(n) = truncate(-sqrt(-7+O(2^(n+2))))\2^n

Formula

a(0) = 1, a(1) = 0; for n >= 2, a(n) = 0 if A318960(n)^2 + 7 is divisible by 2^(n+2), otherwise 1.
a(n) = 1 - A318963(n) for n >= 1.
For n >= 2, a(n) = (A318960(n+1) - A318960(n))/2^n.

Extensions

Corrected by Jianing Song, Aug 28 2019

A318963 Digits of one of the two 2-adic integers sqrt(-7) that ends in 11.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0
Offset: 0

Views

Author

Jianing Song, Sep 06 2018

Keywords

Comments

Over the 2-adic integers there are 2 solutions to x^2 = -7, one ends in 01 and the other ends in 11. This sequence gives the latter one. See A318961 for detailed information.

Examples

			...01001110001100011011001110011111101001011.

Links

Crossrefs

Cf. A318961.
Digits of p-adic integers:
A318962, this sequence (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A212152, A212155 (7-adic, (1+sqrt(-3))/2);
A051277, A290558 (7-adic, sqrt(2));
A286838, A286839 (13-adic, sqrt(-1));
A309989, A309990 (17-adic, sqrt(-1)).
Also there are numerous sequences related to digits of 10-adic integers.

Programs

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

Formula

a(0) = a(1) = 1; for n >= 2, a(n) = 0 if A318961(n)^2 + 7 is divisible by 2^(n+2), otherwise 1.
a(n) = 1 - A318962(n) for n >= 1.
For n >= 2, a(n) = (A318961(n+1) - A318961(n))/2^n.

Extensions

Corrected by Jianing Song, Aug 28 2019

A309989 Digits of one of the two 17-adic integers sqrt(-1).

Original entry on oeis.org

4, 2, 10, 5, 12, 16, 12, 8, 13, 3, 14, 0, 6, 1, 0, 15, 1, 8, 14, 5, 7, 16, 14, 1, 5, 13, 9, 6, 5, 12, 16, 15, 9, 16, 14, 12, 16, 1, 3, 6, 4, 10, 15, 5, 16, 12, 2, 1, 5, 4, 0, 15, 2, 11, 14, 9, 5, 1, 11, 16, 15, 7, 5, 6, 14, 3, 12, 0, 0, 11, 12, 13, 9, 5, 4, 16, 13
Offset: 0

Views

Author

Jianing Song, Aug 26 2019

Keywords

Comments

This square root of -1 in the 17-adic field ends with digit 4. The other, A309990, ends with digit 13 (D when written as a 17-adic number).

Examples

			The solution to x^2 == -1 (mod 17^4) such that x == 4 (mod 17) is x == 27493 (mod 17^4), and 27493 is written as 5A24 in heptadecimal, so the first four terms are 4, 2, 10 and 5.

Links

Crossrefs

Cf. A286877, A286878.
Digits of p-adic square roots:
A318962, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A051277, A290558 (7-adic, sqrt(2));
A321074, A321075 (11-adic, sqrt(3));
A321078, A321079 (11-adic, sqrt(5));
A322091, A322092 (13-adic, sqrt(-3));
A286838, A286839 (13-adic, sqrt(-1));
A322087, A322088 (13-adic, sqrt(3));
this sequence, A309990 (17-adic, sqrt(-1)).

Programs

  • PARI
    a(n) = truncate(sqrt(-1+O(17^(n+1))))\17^n

Formula

a(n) = (A286877(n+1) - A286877(n))/17^n.
For n > 0, a(n) = 16 - A309990(n).

A322561 Digits of one of the two 17-adic integers sqrt(2) that is related to A322559.

Original entry on oeis.org

6, 14, 14, 8, 5, 4, 14, 14, 7, 2, 15, 15, 11, 5, 6, 7, 2, 14, 6, 14, 15, 16, 3, 8, 14, 5, 12, 16, 0, 4, 7, 0, 8, 10, 2, 16, 16, 15, 9, 7, 12, 9, 14, 14, 5, 12, 3, 4, 7, 9, 9, 2, 2, 14, 5, 9, 12, 6, 2, 10, 5, 0, 10, 10, 11, 11, 2, 3, 14, 10, 11, 2, 6, 12, 0, 4
Offset: 0

Views

Author

Jianing Song, Aug 29 2019

Keywords

Comments

This square root of 2 in the 17-adic field ends with digit 6. The other, A322562, ends with digit 11 (B when written as a 17-adic number).

Examples

			The solution to x^2 == 2 (mod 17^4) such that x == 6 (mod 17) is x == 43594 (mod 17^4), and 43594 is written as 8EE6 in heptadecimal, so the first four terms are 6, 14, 14 and 8.

Links

Crossrefs

Cf. A322559, A322560.
Digits of 17-adic square roots:
A309989, A309990 (sqrt(-1));
this sequence, A322562 (sqrt(2));
A322565, A322566 (sqrt(-2)).

Programs

  • PARI
    a(n) = truncate(sqrt(2+O(17^(n+1))))\17^n

Formula

a(n) = (A322559(n+1) - A322559(n))/17^n.
For n > 0, a(n) = 16 - A322562(n).
Equals A309989*A322566 = A309990*A322565.

A322562 Digits of one of the two 17-adic integers sqrt(2) that is related to A322560.

Original entry on oeis.org

11, 2, 2, 8, 11, 12, 2, 2, 9, 14, 1, 1, 5, 11, 10, 9, 14, 2, 10, 2, 1, 0, 13, 8, 2, 11, 4, 0, 16, 12, 9, 16, 8, 6, 14, 0, 0, 1, 7, 9, 4, 7, 2, 2, 11, 4, 13, 12, 9, 7, 7, 14, 14, 2, 11, 7, 4, 10, 14, 6, 11, 16, 6, 6, 5, 5, 14, 13, 2, 6, 5, 14, 10, 4, 16, 12
Offset: 0

Views

Author

Jianing Song, Aug 29 2019

Keywords

Comments

This square root of 2 in the 17-adic field ends with digit 11 (B when written as a 17-adic number). The other, A322561, ends with digit 6.

Examples

			The solution to x^2 == 2 (mod 17^4) such that x == 11 (mod 17) is x == 39927 (mod 17^4), and 39927 is written as 822B in heptadecimal, so the first four terms are 11, 2, 2 and 8.

Links

Crossrefs

Cf. A322559, A322560.
Digits of 17-adic square roots:
A309989, A309990 (sqrt(-1));
A322561, this sequence (sqrt(2));
A322565, A322566 (sqrt(-2)).

Programs

  • PARI
    a(n) = truncate(-sqrt(2+O(17^(n+1))))\17^n

Formula

a(n) = (A322560(n+1) - A322560(n))/17^n.
For n > 0, a(n) = 16 - A322561(n).
Equals A309989*A322565 = A309990*A322566.

A322565 Digits of one of the two 17-adic integers sqrt(-2) that is related to A322563.

Original entry on oeis.org

7, 1, 12, 0, 9, 0, 16, 8, 5, 16, 14, 0, 1, 15, 11, 16, 15, 8, 13, 15, 11, 5, 11, 3, 9, 16, 16, 15, 3, 3, 0, 15, 7, 15, 16, 3, 14, 9, 12, 5, 2, 2, 4, 12, 12, 11, 11, 0, 9, 15, 12, 2, 9, 14, 2, 10, 6, 0, 8, 5, 15, 6, 6, 14, 9, 2, 10, 1, 7, 2, 13, 12, 3, 13, 6, 16
Offset: 0

Views

Author

Jianing Song, Aug 29 2019

Keywords

Comments

This square root of -2 in the 17-adic field ends with digit 7. The other, A322566, ends with digit 10 (A when written as a 17-adic number).

Examples

			The solution to x^2 == -2 (mod 17^4) such that x == 7 (mod 17) is x == 3492 (mod 17^4), and 3492 is written as C17 in heptadecimal, so the first four terms are 7, 1, 12 and 0.

Links

Crossrefs

Cf. A322563, A322564.
Digits of 17-adic square roots:
A309989, A309990 (sqrt(-1));
A322561, A322562 (sqrt(2));
this sequence, A322566 (sqrt(-2)).

Programs

  • PARI
    a(n) = truncate(sqrt(-2+O(17^(n+1))))\17^n

Formula

a(n) = (A322563(n+1) - A322563(n))/17^n.
For n > 0, a(n) = 16 - A322566(n).
Equals A309989*A322561 = A309990*A322562.

A322566 Digits of one of the two 17-adic integers sqrt(-2) that is related to A322564.

Original entry on oeis.org

10, 15, 4, 16, 7, 16, 0, 8, 11, 0, 2, 16, 15, 1, 5, 0, 1, 8, 3, 1, 5, 11, 5, 13, 7, 0, 0, 1, 13, 13, 16, 1, 9, 1, 0, 13, 2, 7, 4, 11, 14, 14, 12, 4, 4, 5, 5, 16, 7, 1, 4, 14, 7, 2, 14, 6, 10, 16, 8, 11, 1, 10, 10, 2, 7, 14, 6, 15, 9, 14, 3, 4, 13, 3, 10, 0
Offset: 0

Views

Author

Jianing Song, Aug 29 2019

Keywords

Comments

This square root of -2 in the 17-adic field ends with digit 10 (A when written as a 17-adic number). The other, A322565, ends with digit 7.

Examples

			The solution to x^2 == -2 (mod 17^4) such that x == 10 (mod 17) is x == 80029 (mod 17^4), and 80029 is written as G4FA in heptadecimal, so the first four terms are 10, 15, 4 and 16.

Links

Crossrefs

Cf. A322563, A322564.
Digits of 17-adic square roots:
A309989, A309990 (sqrt(-1));
A322561, A322562 (sqrt(2));
A322565, this sequence (sqrt(-2)).

Programs

  • PARI
    a(n) = truncate(-sqrt(-2+O(17^(n+1))))\17^n

Formula

a(n) = (A322564(n+1) - A322564(n))/17^n.
For n > 0, a(n) = 16 - A322565(n).
Equals A309989*A322562 = A309990*A322561.
Showing 1-7 of 7 results.

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