A320914 -id:A320914

A321106

Digits of one of the three 13-adic integers 5^(1/3) that is related to A320914.

+20
13

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,1

COMMENTS

For k not divisible by 5, k is a cube in 13-adic field if and only if k == 1, 5, 8, 12 (mod 13). If k is a cube in 13-adic field, then k has exactly three cubic roots.

LINKS

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Wikipedia, p-adic number

FORMULA

a(n) = (A320914(n+1) - A320914(n))/13^n.

EXAMPLE

The unique number k in [1, 13^3] and congruent to 7 modulo 13 such that k^3 - 5 is divisible by 13^3 is k = 1021 = (607)_13, so the first three terms are 7, 0 and 6.

PROG

(PARI) a(n) = lift(sqrtn(5+O(13^(n+1)), 3) * (-1+sqrt(-3+O(13^(n+1))))/2)\13^n

CROSSREFS

Cf. A320914, A320915, A321105, A321107, A321108.

For 5-adic cubic roots, see A290566, A290563, A309443.

KEYWORD

nonn,base

AUTHOR

Jianing Song, Aug 27 2019

STATUS

approved

A321107

Digits of one of the three 13-adic integers 5^(1/3) that is related to A320915.

+10
13

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,1

COMMENTS

For k not divisible by 5, k is a cube in 13-adic field if and only if k == 1, 5, 8, 12 (mod 13). If k is a cube in 13-adic field, then k has exactly three cubic roots.

LINKS

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Wikipedia, p-adic number

FORMULA

a(n) = (A320915(n+1) - A320915(n))/13^n.

EXAMPLE

The unique number k in [1, 13^3] and congruent to 8 modulo 13 such that k^3 - 5 is divisible by 13^3 is k = 177 = (108)_13, so the first three terms are 8, 0 and 1.

PROG

(PARI) a(n) = lift(sqrtn(5+O(13^(n+1)), 3))\13^n

CROSSREFS

Cf. A320914, A320915, A321105, A321106, A321108.

For 5-adic cubic roots, see A290566, A290563, A309443.

KEYWORD

nonn,base

AUTHOR

Jianing Song, Aug 27 2019

STATUS

approved

A321108

Digits of one of the three 13-adic integers 5^(1/3) that is related to A321105.

+10
13

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,1

COMMENTS

For k not divisible by 5, k is a cube in 13-adic field if and only if k == 1, 5, 8, 12 (mod 13). If k is a cube in 13-adic field, then k has exactly three cubic roots.

LINKS

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Wikipedia, p-adic number

FORMULA

a(n) = (A321105(n+1) - A321105(n))/13^n.

EXAMPLE

The unique number k in [1, 13^3] and congruent to 11 modulo 13 such that k^3 - 5 is divisible by 13^3 is k = 999 = (5BB)_13, so the first three terms are 11, 11 and 5.

PROG

(PARI) a(n) = lift(sqrtn(5+O(13^(n+1)), 3) * (-1-sqrt(-3+O(13^(n+1))))/2)\13^n

CROSSREFS

Cf. A320914, A320915, A321105, A321106, A321107.

For 5-adic cubic roots, see A290566, A290563, A309443.

KEYWORD

nonn,base

AUTHOR

Jianing Song, Aug 27 2019

STATUS

approved

A320915

One of the three successive approximations up to 13^n for 13-adic integer 5^(1/3). This is the 8 (mod 13) case (except for n = 0).

+10
12

0, 8, 8, 177, 11162, 211089, 211089, 24345134, 777327338, 7303173106, 113348166836, 1629791577175, 12382753941397, 222065520043726, 1130690839820485, 16880196382617641, 272809661453071426, 5596142534918510154, 14246558454299848087, 576523593214086813732, 4962284464340425145763

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 8 mod 13 such that k^3 - 5 is divisible by 13^n.

For k not divisible by 13, k is a cube in 13-adic field if and only if k == 1, 5, 8, 12 (mod 13). If k is a cube in 13-adic field, then k has exactly three cubic roots.

LINKS

Table of n, a(n) for n=0..20.

Wikipedia, p-adic number

EXAMPLE

The unique number k in [1, 13^2] and congruent to 8 modulo 13 such that k^3 - 5 is divisible by 13^2 is k = 8, so a(2) = 8.

The unique number k in [1, 13^3] and congruent to 8 modulo 13 such that k^3 - 5 is divisible by 13^3 is k = 177, so a(3) = 177.

PROG

(PARI) a(n) = lift(sqrtn(5+O(13^n), 3))

CROSSREFS

Cf. A320914, A321105, A321106, A321107, A321108.

For 5-adic cubic roots, see A290567, A290568, A309444.

KEYWORD

nonn

AUTHOR

Jianing Song, Aug 27 2019

STATUS

approved

A321105

One of the three successive approximations up to 13^n for 13-adic integer 5^(1/3). This is the 11 (mod 13) case (except for n = 0).

+10
12

0, 11, 154, 999, 25166, 82288, 82288, 43523569, 43523569, 4937907895, 121587400998, 1362313827639, 12115276191861, 175201872049228, 2901077831379505, 10775830602778083, 471448867729594896, 6460198350378213465, 23761030189140889331, 361127251045013068718, 4746888122171351400749

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 11 mod 13 such that k^3 - 5 is divisible by 13^n.

For k not divisible by 13, k is a cube in 13-adic field if and only if k == 1, 5, 8, 12 (mod 13). If k is a cube in 13-adic field, then k has exactly three cubic roots.

LINKS

Table of n, a(n) for n=0..20.

Wikipedia, p-adic number

EXAMPLE

The unique number k in [1, 13^2] and congruent to 11 modulo 13 such that k^3 - 5 is divisible by 13^2 is k = 154, so a(2) = 154.

The unique number k in [1, 13^3] and congruent to 11 modulo 13 such that k^3 - 5 is divisible by 13^3 is k = 999, so a(3) = 999.

PROG

(PARI) a(n) = lift(sqrtn(5+O(13^n), 3) * (-1-sqrt(-3+O(13^n)))/2)

CROSSREFS

Cf. A320914, A320915, A321106, A321107, A321108.

For 5-adic cubic roots, see A290567, A290568, A309444.

KEYWORD

nonn

AUTHOR

Jianing Song, Aug 27 2019

STATUS

approved

A319097

One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 3 (mod 7) case (except for n = 0).

+10
11

0, 3, 24, 122, 808, 10412, 111254, 817148, 1640691, 24699895, 186114323, 186114323, 6118094552, 6118094552, 490563146587, 2525232365134, 26263039914849, 59495970484450, 1222648540420485, 6107889334151832, 74501260446390690, 234085793041614692, 1351177521208182706, 24810103812706111000, 134285093173029776372

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 3 mod 7 such that k^3 - 6 is divisible by 7^n.

For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 7). If k is a cube in 7-adic field, then k has exactly three cubic roots.

LINKS

Table of n, a(n) for n=0..24.

Wikipedia, p-adic number

FORMULA

a(n) = A319098(n)*(A210852(n)-1) mod 7^n = A319098(n)*A210852(n)^2 mod 7^n.

a(n) = A319199(n)*(A212153(n)-1) mod 7^n = A319199(n)*A212153(n)^2 mod 7^n.

EXAMPLE

The unique number k in [1, 7^2] and congruent to 3 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 24, so a(2) = 24.

The unique number k in [1, 7^3] and congruent to 3 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 122, so a(3) = 122.

PROG

(PARI) a(n) = lift(sqrtn(6+O(7^n), 3) * (-1+sqrt(-3+O(7^n)))/2)

CROSSREFS

Cf. A319297, A319305, A319555.

Approximations of p-adic cubic roots:

A290567 (5-adic, 2^(1/3));

A290568 (5-adic, 3^(1/3));

A309444 (5-adic, 4^(1/3));

this sequence, A319098, A319199 (7-adic, 6^(1/3));

A320914, A320915, A321105 (13-adic, 5^(1/3)).

KEYWORD

nonn

AUTHOR

Jianing Song, Aug 27 2019

STATUS

approved

A319098

One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 5 (mod 7) case (except for n = 0).

+10
11

0, 5, 40, 138, 824, 3225, 87260, 793154, 793154, 29617159, 191031587, 1320932583, 7252912812, 7252912812, 7252912812, 2041922131359, 16284606661188, 82750467800390, 1013272523749218, 9155340513301463, 31953130884047749, 111745397181659750, 670291261264943757

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 5 mod 7 such that k^3 - 6 is divisible by 7^n.

For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 7). If k is a cube in 7-adic field, then k has exactly three cubic roots.

LINKS

Table of n, a(n) for n=0..22.

Wikipedia, p-adic number

FORMULA

a(n) = A319097(n)*(A212153(n)-1) mod 7^n = A319097(n)*A212153(n)^2 mod 7^n.

a(n) = A319199(n)*(A210852(n)-1) mod 7^n = A319199(n)*A210852(n)^2 mod 7^n.

EXAMPLE

The unique number k in [1, 7^2] and congruent to 5 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 40, so a(2) = 40.

The unique number k in [1, 7^3] and congruent to 5 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 138, so a(3) = 138.

PROG

(PARI) a(n) = lift(sqrtn(6+O(7^n), 3) * (-1-sqrt(-3+O(7^n)))/2)

CROSSREFS

Cf. A319297, A319305, A319555.

Approximations of p-adic cubic roots:

A290567 (5-adic, 2^(1/3));

A290568 (5-adic, 3^(1/3));

A309444 (5-adic, 4^(1/3));

A319097, this sequence, A319199 (7-adic, 6^(1/3));

A320914, A320915, A321105 (13-adic, 5^(1/3)).

KEYWORD

nonn

AUTHOR

Jianing Song, Aug 27 2019

STATUS

approved

A319199

One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 6 (mod 7) case (except for n = 0).

+10
11

0, 6, 34, 83, 769, 3170, 36784, 36784, 3330956, 26390160, 187804588, 470279837, 470279837, 83518003043, 180407013450, 180407013450, 23918214563165, 90384075702367, 1020906131651195, 7534560523292991, 53130141264785563, 212714673860009565, 1888352266109861586

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 6 mod 7 such that k^3 - 6 is divisible by 7^n.

For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 7). If k is a cube in 7-adic field, then k has exactly three cubic roots.

LINKS

Table of n, a(n) for n=0..22.

Wikipedia, p-adic number

FORMULA

a(n) = A319097(n)*(A210852(n)-1) mod 7^n = A319097(n)*A210852(n)^2 mod 7^n.

a(n) = A319098(n)*(A212153(n)-1) mod 7^n = A319098(n)*A212153(n)^2 mod 7^n.

EXAMPLE

The unique number k in [1, 7^2] and congruent to 6 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 34, so a(2) = 24.

The unique number k in [1, 7^3] and congruent to 6 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 83, so a(3) = 122.

PROG

(PARI) a(n) = lift(sqrtn(6+O(7^n), 3))

CROSSREFS

Cf. A319297, A319305, A319555.

Approximations of p-adic cubic roots:

A290567 (5-adic, 2^(1/3));

A290568 (5-adic, 3^(1/3));

A309444 (5-adic, 4^(1/3));

A319097, A319098, this sequence (7-adic, 6^(1/3));

A320914, A320915, A321105 (13-adic, 5^(1/3)).

KEYWORD

nonn

AUTHOR

Jianing Song, Aug 27 2019

STATUS

approved

A322701

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

+10
5

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

COMMENTS

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

LINKS

Table of n, a(n) for n=0..33.

Wikipedia, p-adic number

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).

EXAMPLE

11^3 = 1331 = 83*2^4 + 3;

27^3 = 19683 = 615*2^5 + 3;

59^3 = 205379 = 3209*2^6 + 3.

PROG

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

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)).

KEYWORD

nonn

AUTHOR

Jianing Song, Aug 30 2019

STATUS

approved

A322926

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

+10
5

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,4

COMMENTS

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

LINKS

Table of n, a(n) for n=0..34.

Wikipedia, p-adic number

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).

EXAMPLE

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.

PROG

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

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)).

KEYWORD

nonn

AUTHOR

Jianing Song, Aug 30 2019

STATUS

approved