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.

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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

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Author

Jianing Song, Sep 06 2018

Keywords

Comments

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.

Examples

			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.
...

Links

Crossrefs

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).

Programs

  • PARI
    a(n) = if(n==2, 3, truncate(sqrt(-7+O(2^(n+1)))))

Formula

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.

Extensions

Offset corrected by Jianing Song, Aug 28 2019

A322086 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 9 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 9, 61, 1075, 9863, 9863, 3722793, 56817692, 245063243, 2692255406, 23901254152, 1540344664491, 12293307028713, 198677988008561, 804428201193067, 24428686515388801, 75614579529479558, 741031188712659399, 26692278946856673198, 813880127610558425101, 11047322160238681199840
Offset: 0

Views

Author

Jianing Song, Nov 26 2018

Keywords

Comments

For n > 0, a(n) is the unique solution to x^2 == 3 (mod 13^n) in the range [0, 13^n - 1] and congruent to 9 modulo 13.
A322085 is the approximation (congruent to 4 mod 13) of another square root of 3 over the 13-adic field.

Examples

			9^2 = 81 = 6*13 + 3.
61^2 = 3721 = 22*13^2 + 3.
1075^2 = 1155625 = 526*13^3 + 3.

Links

Crossrefs

Cf. A286840, A286841, A322085, A322088, A322089, A322090.

Programs

  • Maple
    S:= map(t -> op([1,3],t),[padic:-evalp(RootOf(x^2-3,x),13,30)]):
S9:= op(select(t -> t[1]=9, S)):
seq(add(S9[i]*13^(i-1),i=1..n-1),n=1..31); # Robert Israel, Jun 13 2019
  • PARI
    a(n) = truncate(-sqrt(3+O(13^n)))

Formula

For n > 0, a(n) = 13^n - A322085(n).
a(n) = Sum_{i=0..n-1} A322088(i)*13^i.
a(n) = A286840(n)*A322090(n) mod 13^n = A286841(n)*A322089(n) mod 13^n.

A322089 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-3). Here the 6 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 6, 45, 2073, 15255, 300865, 2899916, 22207152, 273201220, 7614777709, 92450772693, 1333177199334, 4917497987408, 191302178967256, 1705677711928521, 48954194340319989, 202511873382592260, 3529594919298491465, 38131258596823843197, 38131258596823843197, 8809653000849500507259
Offset: 0

Views

Author

Jianing Song, Nov 26 2018

Keywords

Comments

For n > 0, a(n) is the unique solution to x^2 == -3 (mod 13^n) in the range [0, 13^n - 1] and congruent to 6 modulo 13.
A322090 is the approximation (congruent to 7 mod 13) of another square root of -3 over the 13-adic field.

Examples

			6^2 = 36 = 3*13 - 3.
45^2 = 2025 = 12*13^2 - 3.
2073^2 = 4297329 = 1956*13^3 - 3.

Links

Crossrefs

Cf. A114525, A286840, A286841, A322085, A322086, A322090, A322091.

Programs

  • PARI
    a(n) = truncate(sqrt(-3+O(13^n)))

Formula

For n > 0, a(n) = 13^n - A322090(n).
a(n) = Sum_{i=0..n-1} A322091(i)*13^i.
a(n) = A286840(n)*A322086(n) mod 13^n = A286841(n)*A322085(n) mod 13^n.
a(n) == L(13^n,6) (mod 13^n) == (3 + sqrt(10))^(13^n) + (3 - sqrt(10))^(13^n) (mod 13^n), where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022

A322090 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 7 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 7, 124, 124, 13306, 70428, 1926893, 40541365, 542529501, 2989721664, 45407719156, 458983194703, 18380587135073, 111572927624997, 2231698673770768, 2231698673770768, 462904735800587581, 5120821000082846468, 74324148355133549932, 1423789031778622267480, 10195310774031298931542
Offset: 0

Views

Author

Jianing Song, Nov 26 2018

Keywords

Comments

For n > 0, a(n) is the unique solution to x^2 == -3 (mod 13^n) in the range [0, 13^n - 1] and congruent to 7 modulo 13.
A322089 is the approximation (congruent to 6 mod 13) of another square root of -3 over the 13-adic field.

Examples

			7^2 = 49 = 4*13 - 3.
124^2 = 15376 = 91*13^2 - 3 = 7*13^3 - 3.
13306^2 = 177049636 = 6199*13^4 - 3.

Links

Crossrefs

Cf. A114525, A286840, A286841, A322085, A322086, A322089, A322092.

Programs

  • PARI
    a(n) = truncate(-sqrt(-3+O(13^n)))

Formula

For n > 0, a(n) = 13^n - A322089(n).
a(n) = Sum_{i=0..n-1} A322092(i)*13^i.
a(n) = A286840(n)*A322085(n) mod 13^n = A286841(n)*A322086(n) mod 13^n.
a(n) == L(13^n,7) (mod 13^n) == ((7 + sqrt(53))/2)^(13^n) + ((7 - sqrt(53))/2)^(13^n) (mod 13^n), where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022

A034944 Successive approximations to 13-adic integer sqrt(-1).

Original entry on oeis.org

0, 5, 70, 239, 143044, 1999509, 6826318, 822557039, 85658552023, 1188526486815, 11941488851037, 291518510320809, 2108769149874327, 13920898306972194, 2675587335039691558, 63228498770709057089
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

References

  • K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973, p. 35.

Links

Crossrefs

Cf. A034935, A034939, A218709, A286840.

Programs

  • PARI
    seq(n)={my(v=vector(n), i=1, k=0); while(i<#v, k++; my(t=truncate(sqrt(-1 + O(13^k)))); if(t > v[i], i++; v[i]=t)); v} \\ Andrew Howroyd, Nov 10 2018
Previous Showing 11-15 of 15 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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