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 1-9 of 9 results.

A325485 One of the four successive approximations up to 5^n for the 5-adic integer 6^(1/4). This is the 2 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 2, 22, 22, 397, 397, 6647, 6647, 319147, 319147, 6178522, 6178522, 103834772, 592116022, 3033522272, 9137037897, 70172194147, 222760084772, 3274517897272, 3274517897272, 60494976881647, 441964703444147, 1395639019850397, 3779824810866022, 51463540631178522
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 2 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 2 modulo 5 such that k^4 - 6 is divisible by 5^2 is k = 22, so a(2) = 22.
The unique number k in [1, 5^3] and congruent to 2 modulo 5 such that k^4 - 6 is divisible by 5^3 is also k = 22, so a(3) is also 22.

Links

Crossrefs

Cf. A048898, A048899, A324024, A325489, A325490, A325491, A325492.
Approximations of p-adic fourth-power roots:
A325484, this sequence, A325486, A325487 (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) * sqrt(-1+O(5^n)))

Formula

a(n) = A325484(n)*A048898(n) mod 13^n = A325485(n)*A048899(n) mod 13^n.
For n > 0, a(n) = 5^n - A325486(n).
a(n)^2 == A324024(n) (mod 5^n).

A325490 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 2 mod 5.

Original entry on oeis.org

2, 4, 0, 3, 0, 2, 0, 4, 0, 3, 0, 2, 2, 2, 1, 2, 1, 4, 0, 3, 4, 2, 1, 4, 1, 1, 2, 0, 0, 3, 0, 1, 1, 3, 1, 4, 4, 0, 2, 4, 0, 4, 1, 2, 0, 1, 2, 3, 2, 4, 2, 4, 1, 3, 0, 2, 1, 0, 3, 3, 3, 3, 0, 2, 2, 3, 1, 1, 4, 1, 1, 0, 1, 4, 0, 3, 3, 3, 0, 3, 0, 0, 4, 0, 3, 2, 3, 1
Offset: 0

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

One of the two square roots of A324026, where an A-number represents a 5-adic number. The other square root is A325491.
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, 5^3] and congruent to 2 modulo 5 such that k^4 - 6 is divisible by 5^3 is k = 22 = (42)_5, so the first three terms are 2, 4 and 0.

Links

Crossrefs

Cf. A210850, A210851, A324026, A325484, A325485, A325486, A325487.
Digits of p-adic fourth-power roots:
A325489, this sequence, A325491, A325492 (5-adic, 6^(1/4));
A324085, A324086, A324087, A324153 (13-adic, 3^(1/4)).

Programs

  • Maple
    S:= select(t -> op([1,3,1],t)=2, [padic:-rootp(_Z^4-6,5,100)]):
op([1,1,3],S); # Robert Israel, Mar 23 2023
  • PARI
    a(n) = lift(sqrtn(6+O(5^(n+1)), 4) * sqrt(-1+O(5^(n+1))))\5^n

Formula

Equals A325489*A210850 = A325492*A210851.
a(n) = (A325485(n+1) - A325485(n))/13^n.
For n > 0, a(n) = 4 - A325491(n).

A325484 One of the four successive approximations up to 5^n for the 5-adic integer 6^(1/4). This is the 1 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 1, 21, 121, 246, 2121, 5246, 52121, 286496, 677121, 677121, 20208371, 117864621, 606145871, 3047552121, 3047552121, 94600286496, 704951848996, 2993770208371, 2993770208371, 79287715520871, 270022578802121, 746859737005246, 5515231319036496, 29357089229192746
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 1 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 1 modulo 5 such that k^4 - 6 is divisible by 5^2 is k = 21, so a(2) = 21.
The unique number k in [1, 5^3] and congruent to 1 modulo 5 such that k^4 - 6 is divisible by 5^3 is k = 121, so a(3) = 121.

Links

Crossrefs

Cf. A048898, A048899, A324023, A325489, A325490, A325491, A325492.
Approximations of p-adic fourth-power roots:
this sequence, A325485, A325486, A325487 (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)*A048899(n) mod 5^n = A325486(n)*A048898(n) mod 5^n.
For n > 0, a(n) = 5^n - A325487(n).
a(n)^2 == A324023(n) (mod 5^n).

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).

A325489 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 1 mod 5.

Original entry on oeis.org

1, 4, 4, 1, 3, 1, 3, 3, 1, 0, 2, 2, 2, 2, 0, 3, 4, 3, 0, 4, 2, 1, 2, 2, 0, 1, 1, 2, 4, 2, 3, 4, 2, 1, 2, 3, 4, 3, 1, 0, 3, 2, 3, 4, 2, 3, 4, 4, 4, 2, 2, 2, 4, 1, 1, 0, 2, 1, 3, 3, 2, 0, 0, 1, 2, 4, 4, 1, 0, 4, 1, 0, 2, 4, 0, 2, 2, 0, 1, 3, 1, 1, 4, 3, 4, 1, 2, 2
Offset: 0

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

One of the two square roots of A324025, where an A-number represents a 5-adic number. The other square root is A325492.
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, 5^3] and congruent to 1 modulo 5 such that k^4 - 6 is divisible by 5^3 is k = 121 = (441)_5, so the first three terms are 1, 4 and 4.

Links

Crossrefs

Cf. A210850, A210851, A324025, A325484, A325485, A325486, A325487.
Digits of p-adic fourth-power roots:
this sequence, A325490, A325491, A325492 (5-adic, 6^(1/4));
A324085, A324086, A324087, A324153 (13-adic, 3^(1/4)).

Programs

  • PARI
    a(n) = lift(sqrtn(6+O(5^(n+1)), 4))\5^n

Formula

Equals A325490*A210851 = A325491*A210850.
a(n) = (A325484(n+1) - A325484(n))/5^n.
For n > 0, a(n) = 4 - A325492(n).

A325491 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 3 mod 5.

Original entry on oeis.org

3, 0, 4, 1, 4, 2, 4, 0, 4, 1, 4, 2, 2, 2, 3, 2, 3, 0, 4, 1, 0, 2, 3, 0, 3, 3, 2, 4, 4, 1, 4, 3, 3, 1, 3, 0, 0, 4, 2, 0, 4, 0, 3, 2, 4, 3, 2, 1, 2, 0, 2, 0, 3, 1, 4, 2, 3, 4, 1, 1, 1, 1, 4, 2, 2, 1, 3, 3, 0, 3, 3, 4, 3, 0, 4, 1, 1, 1, 4, 1, 4, 4, 0, 4, 1, 2, 1, 3
Offset: 0

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

One of the two square roots of A324026, where an A-number represents a 5-adic number. The other square root is A325490.
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, 5^3] and congruent to 3 modulo 5 such that k^4 - 6 is divisible by 5^3 is k = 103 = (403)_5, so the first three terms are 3, 0 and 4.

Links

Crossrefs

Cf. A210850, A210851, A324026, A325484, A325485, A325486, A325487.
Digits of p-adic fourth-power roots:
A325489, A325490, this sequence, A325492 (5-adic, 6^(1/4));
A324085, A324086, A324087, A324153 (13-adic, 3^(1/4)).

Programs

  • PARI
    a(n) = lift(-sqrtn(6+O(5^(n+1)), 4) * sqrt(-1+O(5^(n+1))))\5^n

Formula

Equals A325489*A210851 = A325492*A210850.
a(n) = (A325486(n+1) - A325486(n))/13^n.
For n > 0, a(n) = 4 - A325490(n).

A325492 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 4 mod 5.

Original entry on oeis.org

4, 0, 0, 3, 1, 3, 1, 1, 3, 4, 2, 2, 2, 2, 4, 1, 0, 1, 4, 0, 2, 3, 2, 2, 4, 3, 3, 2, 0, 2, 1, 0, 2, 3, 2, 1, 0, 1, 3, 4, 1, 2, 1, 0, 2, 1, 0, 0, 0, 2, 2, 2, 0, 3, 3, 4, 2, 3, 1, 1, 2, 4, 4, 3, 2, 0, 0, 3, 4, 0, 3, 4, 2, 0, 4, 2, 2, 4, 3, 1, 3, 3, 0, 1, 0, 3, 2, 2
Offset: 0

Views

Author

Jianing Song, Sep 07 2019

Keywords

Comments

One of the two square roots of A324025, where an A-number represents a 5-adic number. The other square root is A325489.
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, 5^3] and congruent to 4 modulo 5 such that k^4 - 6 is divisible by 5^3 is k = 4 = (4)_5, so the first three terms are 4, 0 and 0.

Links

Crossrefs

Cf. A210850, A210851, A324025, A325484, A325485, A325486, A325487.
Digits of p-adic fourth-power roots:
A325489, A325490, A325491, this sequence (5-adic, 6^(1/4));
A324085, A324086, A324087, A324153 (13-adic, 3^(1/4)).

Programs

  • PARI
    a(n) = lift(-sqrtn(6+O(5^(n+1)), 4))\5^n

Formula

Equals A325490*A210850 = A325491*A210851.
a(n) = (A325487(n+1) - A325487(n))/5^n.
For n > 0, a(n) = 4 - A325489(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).
Showing 1-9 of 9 results.

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