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.

Previous Showing 11-13 of 13 results.

A325487 One of the four successive approximations up to 13^n for the 13-adic integer 6^(1/4). This is the 4 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 4, 4, 4, 379, 1004, 10379, 26004, 104129, 1276004, 9088504, 28619754, 126276004, 614557254, 3055963504, 27470026004, 57987604129, 57987604129, 820927057254, 16079716119754, 16079716119754, 206814579401004, 1637326054010379, 6405697636041629, 30247555546197879
Offset: 0

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

For n > 0, a(n) is the unique number k in [1, 5^n] and congruent to 4 mod 5 such that k^4 - 6 is divisible by 5^n.
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, 5^2] and congruent to 4 modulo 5 such that k^4 - 6 is divisible by 5^2 is k = 4, so a(2) = 4.
The unique number k in [1, 5^3] and congruent to 4 modulo 5 such that k^4 - 6 is divisible by 5^3 is also k = 4, so a(3) is also 4.

Links

Crossrefs

Cf. A048898, A048899, A324023, A325489, A325490, A325491, A325492.
Approximations of p-adic fourth-power roots:
A325484, A325485, A325486, this sequence (5-adic, 6^(1/4));
A324077, A324082, A324083, A324084 (13-adic, 3^(1/4)).

Programs

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

Formula

a(n) = A325485(n)*A048898(n) mod 5^n = A325486(n)*A048899(n) mod 5^n.
For n > 0, a(n) = 5^n - A325484(n).
a(n)^2 == A324023(n) (mod 5^n).

A341751 Successive approximations up to 2^n for the 2-adic integer 17^(1/4). This is the 1 (mod 4) case.

Original entry on oeis.org

1, 5, 13, 13, 45, 45, 173, 429, 429, 1453, 3501, 7597, 7597, 23981, 23981, 23981, 155053, 417197, 941485, 1990061, 1990061, 1990061, 10378669, 10378669, 10378669, 10378669, 144596397, 413031853, 413031853, 413031853, 2560515501, 6855482797, 15445417389, 15445417389
Offset: 2

Views

Author

Jianing Song, Feb 18 2021

Keywords

Comments

a(n) is the unique number k in [1, 2^n] and congruent to 1 mod 4 such that k^4 - 17 is divisible by 2^(n+2).
For odd k, k has a fourth root in the ring of 2-adic integers if and only if k == 1 (mod 16), in which case k has exactly two fourth roots.

Examples

			The unique number k in [1, 4] and congruent to 1 modulo 4 such that k^4 - 17 is divisible by 16 is 1, so a(2) = 1.
a(2)^4 - 17 = -16 which is not divisible by 32, so a(3) = a(2) + 2^2 = 5.
a(3)^4 - 17 = 608 which is not divisible by 64, so a(4) = a(3) + 2^3 = 13.
a(4)^4 - 17 = 28544 which is divisible by 128, so a(5) = a(4) = 13.
a(5)^4 - 17 = 28544 which is not ndivisible by 256, so a(6) = a(5) + 2^5 = 45.
...

Links

Crossrefs

Cf. A341753 (digits of the associated 2-adic fourth root of 17), A341538.
Approximations of p-adic fourth-power roots:
this sequence, A341752 (2-adic, 17^(1/4));
A325484, A325485, A325486, A325487 (5-adic, 6^(1/4));
A324077, A324082, A324083, A324084 (13-adic, 3^(1/4)).

Programs

  • PARI
    a(n) = truncate(sqrtn(17+O(2^(n+2)), 4))

Formula

a(2) = 1; for n >= 3, a(n) = a(n-1) if a(n-1)^4 - 17 is divisible by 2^(n+2), otherwise a(n-1) + 2^(n-1).
a(n) = 2^n - A341752(n).
a(n) = Sum_{i=0..n-1} A341753(i)*2^i.
a(n)^2 == A341538(n) (mod 2^n).

A341752 Successive approximations up to 2^n for the 2-adic integer 17^(1/4). This is the 3 (mod 4) case.

Original entry on oeis.org

3, 3, 3, 19, 19, 83, 83, 83, 595, 595, 595, 595, 8787, 8787, 41555, 107091, 107091, 107091, 107091, 107091, 2204243, 6398547, 6398547, 23175763, 56730195, 123839059, 123839059, 123839059, 660709971, 1734451795, 1734451795, 1734451795, 1734451795, 18914320979
Offset: 2

Views

Author

Jianing Song, Feb 18 2021

Keywords

Comments

a(n) is the unique number k in [1, 2^n] and congruent to 1 mod 4 such that k^4 - 17 is divisible by 2^(n+2).
For odd k, k has a fourth root in the ring of 2-adic integers if and only if k == 1 (mod 16), in which case k has exactly two fourth roots.

Examples

			The unique number k in [1, 4] and congruent to 3 modulo 4 such that k^4 - 17 is divisible by 16 is 3, so a(2) = 3.
a(2)^4 - 17 = 64 which is divisible by 32, so a(3) = a(2) = 3.
a(3)^4 - 17 = 64 which is divisible by 64, so a(4) = a(3) = 3.
a(4)^4 - 17 = 64 which is not divisible by 128, so a(5) = a(4) + 2^4 = 19.
a(5)^4 - 17 = 130304 which is ndivisible by 256, so a(6) = a(5) = 19.
...

Links

Crossrefs

Cf. A341754 (digits of the associated 2-adic fourth root of 17), A341538.
Approximations of p-adic fourth-power roots:
A341751, this sequence (2-adic, 17^(1/4));
A325484, A325485, A325486, A325487 (5-adic, 6^(1/4));
A324077, A324082, A324083, A324084 (13-adic, 3^(1/4)).

Programs

  • PARI
    a(n) = truncate(-sqrtn(17+O(2^(n+2)), 4))

Formula

a(2) = 3; for n >= 3, a(n) = a(n-1) if a(n-1)^4 - 17 is divisible by 2^(n+2), otherwise a(n-1) + 2^(n-1).
a(n) = 2^n - A341751(n).
a(n) = Sum_{i=0..n-1} A341754(i)*2^i.
a(n)^2 == A341538(n) (mod 2^n).
Previous Showing 11-13 of 13 results.

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

Content is available under The OEIS End-User License Agreement.