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 1-5 of 5 results.

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.

A322563 One of the two successive approximations up to 17^n for 17-adic integer sqrt(-2). This is the 7 (mod 17) case (except for n = 0).

Original entry on oeis.org

0, 7, 24, 3492, 3492, 755181, 755181, 386956285, 3669665669, 38548452874, 1935954476826, 30159869083112, 30159869083112, 612782106312873, 149181452599901928, 2001337544753312147, 47800106368910364835, 777717984503913392050, 7395640079594607505466
Offset: 0

Views

Author

Jianing Song, Aug 29 2019

Keywords

Comments

For n > 0, a(n) is the unique solution to x^2 == -2 (mod 17^n) in the range [0, 17^n - 1] and congruent to 7 modulo 17.
A322564 is the approximation (congruent to 10 mod 17) of another square root of -2 over the 17-adic field.

Examples

			7^2 = 49 = 3*17 - 2;
24^2 = 576 = 2*17^2 - 2;
3492^2 = 12194064 = 2482*17^3 - 2.

Links

Crossrefs

Cf. A322565, A322566.
Approximations of 17-adic square roots:
A286877, A286878 (sqrt(-1));
A322559, A322560 (sqrt(2));
this sequence, A322564 (sqrt(-2)).

Programs

  • Mathematica
    {0}~Join~Table[First@Select[PowerModList[-2,1/2,17^k],Mod[#,17]==7&],{k,20}] (* Giorgos Kalogeropoulos, Sep 14 2022 *)
  • PARI
    a(n) = truncate(sqrt(-2+O(17^n)))

Formula

For n > 0, a(n) = 17^n - A322564(n).
a(n) = Sum_{i=0..n-1} A322565(i)*17^i.
a(n) = A286877(n)*A322559(n) mod 17^n = A286878(n)*A322560(n) mod 17^n.

A322564 One of the two successive approximations up to 17^n for 17-adic integer sqrt(-2). This is the 10 (mod 17) case (except for n = 0).

Original entry on oeis.org

0, 10, 265, 1421, 80029, 664676, 23382388, 23382388, 3306091772, 80039423623, 80039423623, 4112027224521, 552462368146649, 9291795926593064, 19196373959499001, 861085506756503646, 861085506756503646, 49522277382423372127, 6667444372473117485543, 48856697728676292458570, 287929133413827617305723
Offset: 0

Views

Author

Jianing Song, Aug 29 2019

Keywords

Comments

For n > 0, a(n) is the unique solution to x^2 == 2 (mod 17^n) in the range [0, 17^n - 1] and congruent to 10 modulo 17.
A322563 is the approximation (congruent to 7 mod 17) of another square root of -2 over the 17-adic field.

Examples

			10^2 = 100 = 6*17 - 2;
265^2 = 70225 = 243*17^2 - 2;
1421^2 = 2019241 = 411*17^3 - 2.

Links

Crossrefs

Cf. A322565, A322566.
Approximations of 17-adic square roots:
A286877, A286878 (sqrt(-1));
A322559, A322560 (sqrt(2));
A322563, this sequence (sqrt(-2)).

Programs

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

Formula

For n > 0, a(n) = 17^n - A322563(n).
a(n) = Sum_{i=0..n-1} A322566(i)*17^i.
a(n) = A286877(n)*A322560(n) mod 17^n = A286878(n)*A322559(n) mod 17^n.
Showing 1-5 of 5 results.

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