A324086 -id:A324086 - cp's OEIS frontend

A324153 Digits of one of the four 13-adic integers 3^(1/4) that is congruent to 11 mod 13.

11, 10, 5, 11, 0, 6, 0, 8, 4, 6, 11, 2, 8, 6, 5, 4, 2, 11, 0, 3, 3, 5, 12, 0, 9, 6, 8, 7, 1, 0, 9, 1, 3, 7, 4, 8, 8, 10, 5, 8, 1, 4, 8, 2, 11, 12, 10, 11, 8, 9, 1, 5, 9, 6, 9, 10, 6, 5, 9, 6, 11, 12, 9, 12, 1, 4, 1, 6, 1, 12, 9, 7, 8, 5, 3, 2, 0, 6, 1, 7, 11

Offset: 0

Jianing Song, Sep 01 2019

One of the two square roots of A322087, where an A-number represents a 13-adic number. The other square root is A324085.

For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.

			The unique number k in [1, 13^3] and congruent to 11 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 986 = (5AB)_13, so the first three terms are 11, 10 and 5.

Robert Israel, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number

Cf. A286838, A286839, A322087, A324077, A324082, A324083, A324084, A324085, A324086, A324087.

R:= select(t -> op([1,3,1],t)=11, [padic:-rootp(x^4-3, 13,101)]):
op([1,1,3],R); # Robert Israel, Sep 08 2019

a(n) = lift(-sqrtn(3+O(13^(n+1)), 4) * sqrt(-1+O(13^(n+1))))\13^n

Equals A324086*A286839 = A324087*A286838.
a(n) = (A324084(n+1) - A324084(n))/13^n.
For n > 0, a(n) = 12 - A324085(n).

A324077 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 2 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 2, 28, 1211, 3408, 346140, 2573898, 60495606, 311489674, 6837335442, 70464331680, 208322823529, 18129926763899, 111322267253823, 1928572906807341, 29490207606702364, 438977351719428420, 7093143443551226830, 15743559362932564763, 1365208442786421282311

Offset: 0

Jianing Song, Sep 01 2019

nonn

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 2 mod 13 such that k^4 - 3 is divisible by 13^n.

For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.

			The unique number k in [1, 13^2] and congruent to 2 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 28, so a(2) = 28.
The unique number k in [1, 13^3] and congruent to 2 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 1211, so a(3) = 1211.

Wikipedia, p-adic number

Cf. A286840, A286841, A322085, A324082, A324083, A324084, A324085, A324086, A324087, A324153.

a(n) = lift(sqrtn(3+O(13^n), 4) * sqrt(-1+O(13^n)))

a(n) = A324082(n)*A286840(n) mod 13^n = A324083(n)*A286841(n) mod 13^n.
For n > 0, a(n) = 13^n - A324084(n).
a(n)^2 == A322085(n) (mod 13^n).

A324082 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 3 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 3, 68, 575, 13757, 156562, 4612078, 52880168, 178377202, 9967145854, 137221138330, 1240089073122, 22746013801566, 279024950148857, 2399150696294628, 2399150696294628, 104770936724476142, 3431853982640375347, 98586429095835092610, 1335595905567366417029

Offset: 0

Jianing Song, Sep 01 2019

nonn

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 3 mod 13 such that k^4 - 3 is divisible by 13^n.

For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.

			The unique number k in [1, 13^2] and congruent to 3 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 68, so a(2) = 68.
The unique number k in [1, 13^3] and congruent to 3 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 575, so a(3) = 575.

Wikipedia, p-adic number

Cf. A286840, A286841, A322085, A324077, A324083, A324084, A324085, A324086, A324087, A324153.

PARI
```
a(n) = lift(sqrtn(3+O(13^n), 4))
```

a(n) = A324077(n)*A286841(n) mod 13^n = A324084(n)*A286840(n) mod 13^n.
For n > 0, a(n) = 13^n - A324083(n).
a(n)^2 == A322086(n) (mod 13^n).

A324083 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 10 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 10, 101, 1622, 14804, 214731, 214731, 9868349, 637353519, 637353519, 637353519, 552071320915, 552071320915, 23850156443396, 1538225689404661, 48786742317796129, 560645672458703699, 5218561936740962586, 13868977856122300519, 126324384808079693648

Offset: 0

Jianing Song, Sep 01 2019

nonn

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 10 mod 13 such that k^4 - 3 is divisible by 13^n.

For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.

			The unique number k in [1, 13^2] and congruent to 10 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 101, so a(2) = 101.
The unique number k in [1, 13^3] and congruent to 10 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 1622, so a(3) = 1622.

Wikipedia, p-adic number

Cf. A286840, A286841, A322085, A324077, A324082, A324084, A324085, A324086, A324087, A324153.

PARI
```
a(n) = lift(-sqrtn(3+O(13^n), 4))
```

a(n) = A324077(n)*A286840(n) mod 13^n = A324084(n)*A286841(n) mod 13^n.
For n > 0, a(n) = 13^n - A324082(n).
a(n)^2 == A322086(n) (mod 13^n).

A324084 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 11 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 11, 141, 986, 25153, 25153, 2252911, 2252911, 504241047, 3767163931, 67394160169, 1583837570508, 5168158358582, 191552839338430, 2008803478891948, 21695685407388393, 226439257463751421, 1557272475830111103, 96711847589024828366, 96711847589024828366

Offset: 0

Jianing Song, Sep 01 2019

nonn

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 11 mod 13 such that k^4 - 3 is divisible by 13^n.

For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.

			The unique number k in [1, 13^2] and congruent to 11 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 141, so a(2) = 141.
The unique number k in [1, 13^3] and congruent to 11 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 986, so a(3) = 986.

Wikipedia, p-adic number

Cf. A286840, A286841, A322085, A324077, A324082, A324083, A324085, A324086, A324087, A324153.

a(n) = lift(-sqrtn(3+O(13^n), 4) * sqrt(-1+O(13^n)))

a(n) = A324082(n)*A286841(n) mod 13^n = A324083(n)*A286840(n) mod 13^n.
For n > 0, a(n) = 13^n - A324077(n).
a(n)^2 == A322085(n) (mod 13^n).

A324085 Digits of one of the four 3-adic integers 3^(1/4) that is congruent to 2 mod 13.

Original entry on oeis.org

2, 2, 7, 1, 12, 6, 12, 4, 8, 6, 1, 10, 4, 6, 7, 8, 10, 1, 12, 9, 9, 7, 0, 12, 3, 6, 4, 5, 11, 12, 3, 11, 9, 5, 8, 4, 4, 2, 7, 4, 11, 8, 4, 10, 1, 0, 2, 1, 4, 3, 11, 7, 3, 6, 3, 2, 6, 7, 3, 6, 1, 0, 3, 0, 11, 8, 11, 6, 11, 0, 3, 5, 4, 7, 9, 10, 12, 6, 11, 5, 1

Offset: 0

Jianing Song, Sep 01 2019

One of the two square roots of A322087, where an A-number represents a 13-adic number. The other square root is A324153.

For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.

			The unique number k in [1, 13^3] and congruent to 2 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 1211 = (722)_13, so the first three terms are 2, 7 and 7.

Wikipedia, p-adic number

Cf. A286838, A286839, A322087, A324077, A324082, A324083, A324084, A324086, A324087, A324153.

a(n) = lift(sqrtn(3+O(13^(n+1)), 4) * sqrt(-1+O(13^(n+1))))\13^n

Equals A324086*A286838 = A324087*A286839.
a(n) = (A324077(n+1) - A324077(n))/13^n.
For n > 0, a(n) = 12 - A324153(n).

A324087 Digits of one of the four 3-adic integers 3^(1/4) that is congruent to 10 mod 13.

Original entry on oeis.org

10, 7, 9, 6, 7, 0, 2, 10, 0, 0, 4, 0, 1, 5, 12, 10, 7, 1, 1, 9, 7, 1, 7, 8, 0, 0, 9, 10, 5, 5, 0, 1, 4, 7, 0, 9, 7, 4, 6, 0, 3, 8, 12, 7, 7, 0, 11, 3, 11, 3, 1, 5, 8, 12, 9, 3, 12, 0, 6, 6, 11, 4, 8, 3, 7, 6, 3, 7, 5, 2, 11, 9, 9, 4, 7, 1, 4, 10, 12, 11, 0

Offset: 0

Jianing Song, Sep 01 2019

One of the two square roots of A322088, where an A-number represents a 13-adic number. The other square root is A324086.

For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.

			The unique number k in [1, 13^3] and congruent to 10 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 1622 = (97A)_13, so the first three terms are 10, 7 and 9.

Wikipedia, p-adic number

Cf. A286838, A286839, A322087, A324077, A324082, A324083, A324084, A324085, A324086, A324153.

a(n) = lift(-sqrtn(3+O(13^(n+1)), 4))\13^n

Equals A324085*A286838 = A324153*A286839.
a(n) = (A324083(n+1) - A324083(n))/13^n.
For n > 0, a(n) = 12 - A324086(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

Jianing Song, Sep 07 2019

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.

			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.

Robert Israel, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number

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)).

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

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

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

Jianing Song, Sep 07 2019

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.

			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.

Wikipedia, p-adic number

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)).

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

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

Jianing Song, Sep 07 2019

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.

			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.

Wikipedia, p-adic number

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)).

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

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

cp's OEIS Frontend

A324153 Digits of one of the four 13-adic integers 3^(1/4) that is congruent to 11 mod 13.

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A324077 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 2 (mod 13) case (except for n = 0).

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A324082 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 3 (mod 13) case (except for n = 0).

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A324083 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 10 (mod 13) case (except for n = 0).

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A324084 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 11 (mod 13) case (except for n = 0).

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A324085 Digits of one of the four 3-adic integers 3^(1/4) that is congruent to 2 mod 13.

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A324087 Digits of one of the four 3-adic integers 3^(1/4) that is congruent to 10 mod 13.

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