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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A318960 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 1 (mod 4) case.

Original entry on oeis.org

1, 5, 5, 21, 53, 53, 181, 181, 181, 181, 181, 181, 181, 16565, 49333, 49333, 49333, 49333, 573621, 1622197, 1622197, 1622197, 10010805, 10010805, 10010805, 77119669, 211337397, 479772853, 479772853, 479772853, 2627256501, 6922223797, 15512158389, 15512158389
Offset: 2

Views

Author

Jianing Song, Sep 06 2018

Keywords

Comments

a(n) is the unique number k in [1, 2^n] and congruent to 1 (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 1 modulo 4 such that k^2 + 7 is divisible by 8 is 1, so a(2) = 1.
a(2)^2 + 7 = 8 which is not divisible by 16, so a(3) = a(2) + 2^2 = 5.
a(3)^2 + 7 = 32 which is divisible by 32, so a(4) = a(3) = 5.
a(4)^2 + 7 = 32 which is divisible by 64, so a(5) = a(4) + 2^4 = 21.
a(5)^2 + 7 = 448 which is divisible by 128, so a(6) = a(5) + 2^5 = 53.
...

Links

Crossrefs

Cf. A318962.
Expansions of p-adic integers:
this sequence, A318961 (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) = truncate(-sqrt(-7+O(2^(n+1))))

Formula

a(2) = 1; 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 - A318961(n).
a(n) = Sum_{i=0..n-1} A318962(i)*2^i.

Extensions

Offset corrected by Jianing Song, Aug 28 2019

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

Views

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

A322701 The successive approximations up to 2^n for 2-adic integer 3^(1/3).

Original entry on oeis.org

0, 1, 3, 3, 11, 27, 59, 123, 123, 379, 379, 379, 379, 4475, 12667, 29051, 61819, 127355, 127355, 127355, 127355, 127355, 2224507, 2224507, 2224507, 19001723, 52556155, 119665019, 253882747, 253882747, 253882747, 1327624571, 3475108219, 7770075515
Offset: 0

Views

Author

Jianing Song, Aug 30 2019

Keywords

Comments

a(n) is the unique solution to x^3 == 3 (mod 2^n) in the range [0, 2^n - 1].

Examples

			11^3 = 1331 = 83*2^4 + 3;
27^3 = 19683 = 615*2^5 + 3;
59^3 = 205379 = 3209*2^6 + 3.

Links

Crossrefs

For the digits of 3^(1/3), see A323000.
Approximations of p-adic cubic roots:
this sequence (2-adic, 3^(1/3));
A322926 (2-adic, 5^(1/3));
A322934 (2-adic, 7^(1/3));
A322999 (2-adic, 9^(1/3));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A309444 (5-adic, 4^(1/3));
A319097, A319098, A319199 (7-adic, 6^(1/3));
A320914, A320915, A321105 (13-adic, 5^(1/3)).

Programs

  • PARI
    a(n) = lift(sqrtn(3+O(2^n), 3))

Formula

For n > 0, a(n) = a(n-1) if a(n-1)^3 - 3 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).

A322926 The successive approximations up to 2^n for 2-adic integer 5^(1/3).

Original entry on oeis.org

0, 1, 1, 5, 13, 29, 29, 93, 93, 93, 605, 1629, 3677, 3677, 3677, 20061, 20061, 20061, 151133, 151133, 151133, 151133, 151133, 4345437, 4345437, 21122653, 54677085, 54677085, 188894813, 457330269, 457330269, 457330269, 2604813917, 6899781213, 6899781213
Offset: 0

Views

Author

Jianing Song, Aug 30 2019

Keywords

Comments

a(n) is the unique solution to x^3 == 5 (mod 2^n) in the range [0, 2^n - 1].

Examples

			13^3 = 2197 = 137*2^4 + 5;
29^3 = 24389 = 762*2^5 + 5 = 381*2^6 + 5;
93^3 = 804357 = 6284*2^7 + 5 = 3142*2^8 + 5 = 1571*2^9 + 5.

Links

Crossrefs

For the digits of 5^(1/3), see A323045.
Approximations of p-adic cubic roots:
A322701 (2-adic, 3^(1/3));
this sequence (2-adic, 5^(1/3));
A322934 (2-adic, 7^(1/3));
A322999 (2-adic, 9^(1/3));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A309444 (5-adic, 4^(1/3));
A319097, A319098, A319199 (7-adic, 6^(1/3));
A320914, A320915, A321105 (13-adic, 5^(1/3)).

Programs

  • PARI
    a(n) = lift(sqrtn(5+O(2^n), 3))

Formula

For n > 0, a(n) = a(n-1) if a(n-1)^3 - 5 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).

A322934 The successive approximations up to 2^n for 2-adic integer 7^(1/3).

Original entry on oeis.org

0, 1, 3, 7, 7, 23, 23, 23, 151, 407, 407, 1431, 3479, 3479, 11671, 11671, 44439, 109975, 241047, 503191, 1027479, 2076055, 2076055, 6270359, 6270359, 6270359, 6270359, 6270359, 6270359, 274705815, 811576727, 1885318551, 1885318551, 6180285847
Offset: 0

Views

Author

Jianing Song, Aug 30 2019

Keywords

Comments

a(n) is the unique solution to x^3 == 7 (mod 2^n) in the range [0, 2^n - 1].

Examples

			7^3 = 343 = 21*2^4 + 7;
23^3 = 12167 = 380*2^5 + 7 = 190*2^6 + 7 = 95*2^7 + 7;
151^3 = 3442951 = 13449*2^8 + 7.

Links

Crossrefs

For the digits of 7^(1/3), see A323095.
Approximations of p-adic cubic roots:
A322701 (2-adic, 3^(1/3));
A322926 (2-adic, 5^(1/3));
this sequence (2-adic, 7^(1/3));
A322999 (2-adic, 9^(1/3));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A309444 (5-adic, 4^(1/3));
A319097, A319098, A319199 (7-adic, 6^(1/3));
A320914, A320915, A321105 (13-adic, 5^(1/3)).

Programs

  • PARI
    a(n) = lift(sqrtn(7+O(2^n), 3))

Formula

For n > 0, a(n) = a(n-1) if a(n-1)^3 - 7 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).

A322999 The successive approximations up to 2^n for 2-adic integer 9^(1/3).

Original entry on oeis.org

0, 1, 1, 1, 9, 25, 25, 25, 25, 281, 281, 281, 281, 4377, 4377, 20761, 53529, 53529, 184601, 446745, 971033, 2019609, 4116761, 8311065, 8311065, 25088281, 58642713, 125751577, 259969305, 259969305, 259969305, 259969305, 259969305, 4554936601, 13144871193
Offset: 0

Views

Author

Jianing Song, Aug 30 2019

Keywords

Comments

a(n) is the unique solution to x^3 == 9 (mod 2^n) in the range [0, 2^n - 1].

Examples

			9^3 = 729 = 45*2^4 + 9;
25^3 = 15625 = 488*2^5 + 9 = 244*2^6 + 9 = 122*2^7 + 9 = 61*2^8 + 9;
281^3 = 22188041 = 43336*2^9 + 9 = 21668*2^10 + 9 = 10834*2^11 + 9 = 5417*2^12 + 9.

Links

Crossrefs

For the digits of 9^(1/3), see A323096.
Approximations of p-adic cubic roots:
A322701 (2-adic, 3^(1/3));
A322926 (2-adic, 5^(1/3));
A322934 (2-adic, 7^(1/3));
this sequence (2-adic, 9^(1/3));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A309444 (5-adic, 4^(1/3));
A319097, A319098, A319199 (7-adic, 6^(1/3));
A320914, A320915, A321105 (13-adic, 5^(1/3)).

Programs

  • PARI
    a(n) = lift(sqrtn(9+O(2^n), 3))

Formula

For n > 0, a(n) = a(n-1) if a(n-1)^3 - 9 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).
Previous Showing 11-16 of 16 results.

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