A319098 -id:A319098 - cp's OEIS frontend

A319305 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319098.

5, 5, 2, 2, 1, 5, 6, 0, 5, 4, 4, 3, 0, 0, 3, 3, 2, 4, 5, 2, 1, 1, 6, 1, 4, 6, 2, 2, 2, 6, 2, 4, 1, 0, 0, 2, 0, 3, 4, 2, 2, 1, 5, 2, 6, 1, 0, 4, 1, 6, 0, 6, 0, 1, 3, 0, 3, 4, 1, 1, 1, 0, 6, 5, 2, 4, 0, 2, 2, 5, 1, 1, 0, 4, 0, 0, 5, 6, 4, 1, 5, 2, 4, 3, 0, 1, 2, 5

Offset: 0

Jianing Song, Aug 27 2019

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. [Typo corrected by Keyang Li, Nov 04 2024]

			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 = (255)_7, so the first three terms are 5, 5 and 2.

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

Cf. A319097, A319098, A319199.
Digits of p-adic cubic roots:
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A309443 (5-adic, 4^(1/3));
A319297, this sequence, A319555 (7-adic, 6^(1/3));
A321106, A321107, A321108 (13-adic, 5^(1/3)).

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

Equals A319297*(A212155-1) = A319297*A212155^2, where each A-number represents a 7-adic number.

Equals A319555*(A212152-1) = A319555*A212152^2.

A210852 Approximations up to 7^n for one of the three 7-adic integers (-1)^(1/3).

Original entry on oeis.org

0, 3, 31, 325, 1354, 1354, 34968, 740862, 2387948, 25447152, 146507973, 1276408969, 9185715941, 78392151946, 272170172760, 950393245609, 10445516265495, 43678446835096, 974200502783924, 10744682090246618, 22143577275619761

Offset: 0

Wolfdieter Lang, May 02 2012

The numbers are computed from the recurrence given below in the formula field. This recurrence follows from the formula a(n) = 3^(7^(n-1)) (mod 7^n), n >= 1, which satisfies a(n)^3 + 1 == 0 (mod 7^n), n >= 1. a(0) = 0 satisfies this congruence as well. The proof can be done by showing that each term in the binomial expansion of (3^(7^(n-1)))^3 +1 = (28 -1)^(7^(n-1)) + 1 has a factor 7^n.
a(n) == 3 (mod 7), n >= 1. This follows from the formula given above, and 3^(7^(n-1)) == 3 (mod 7), n >= 1 (proof by induction).
The digit t(n), n >= 0, multiplying 7^n in the 7-adic integer (-1)^(1/3) corresponding to this sequence is obtained from the (unique) solution of the linear congruence 3*a(n)^2*t(n) + b(n) == 0 (mod 7), n >= 1, with b(n):= (a(n)^3+1)/7^n = A210853(n). t(0):=3, one of the three solutions of X^3 + 1 == 0 (mod 7). For these digits see A212152. The 7-adic number is, read from right to left, ...3143214516604202226653431432053116412125443426203643 =: u.
  a(n) is obtained from reading u in base 7, and adding the first n terms.
One can show directly that a(n) = 7^n + 1 - y(n), n >= 1, with y(n) = A212153(n) and z(n) = 7^n - 1 = 6*A023000(n), n >= 0.
Iff a(n+1) = a(n) then t(n) = A212152(n) = 0.
See the Nagell reference given in A210848 for theorems 50 and 52 on p. 87, and formula (6) on page 86, adapted to this case. Because X^3 + 1 = 0 (mod 7) has the three simple roots 3, 5 and 6, one has for X(n)^3 + 1 == 0 (mod 7^n) exactly three solutions for each n >= 1, which can be chosen as a(n) == 3 (mod 7), y(n) == 5 (mod 7) and z(n) == 6 (mod 7) == -1 (mod 7). The y- and z- sequences are given in A212153 and 6*A023000, respectively.
For n > 0, a(n) - 1 (== a(n)^2 (mod 7^n)) and 7^n - a(n) (== a(n)^4 (mod 7^n)) are the two primitive cubic roots of unity in Z/(7^n Z). - Álvar Ibeas, Feb 20 2017
From Jianing Song, Aug 26 2022: (Start)
a(n) is the solution to x^2 - x + 1 == 0 (mod 7^n) that is congruent to 3 modulo 7 (if n>0).
A212153(n) is the multiplicative inverse of a(n) modulo 7^n. (End)

			a(3) == 31^7 (mod 7^3) == 27512614111 (mod 343) = 325.
a(3) == 3^49 (mod 7^3) = 325.
a(3) = 31 + 6*7^2 = 325.
a(3) = 3*7^0 + 4*7^1 + 6*7^2 = 325.
a(3) = 7^3 +1 - 19 = 325.
a(5) = a(4) = 1354 because A212152(4) = 0.

Kenny Lau, Table of n, a(n) for n = 0..1183

Cf. A212152 (digits of (-1)^(1/3)), A212153 (approximations of another cube root of -1), 6*A023000 (approximations of -1).

Cf. A048898, A048899 (approximations of the 5-adic integers sqrt(-1)); A319097, A319098, A319199 (approximations of the 7-adic integers 6^(1/3)).

a:=proc(n) option remember: if n=0 then 0 elif n=1 then 3
else modp(a(n-1)^7, 7^n) fi end proc: [seq(a(n),n=0..30)];

a[n_] := a[n] = Which[n == 0, 0, n == 1, 3, True, Mod[a[n-1]^7, 7^n]]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Mar 05 2014, after Maple *)

a(n) = lift((1-sqrt(-3+O(7^n)))/2) \\ Jianing Song, Aug 26 2022

Recurrence: a(n) = a(n-1)^7 (mod 7^n), n >= 2, a(0)=0, a(1)=3.
a(n) == 3^(7^(n-1)) (mod 7^n) == 3 (mod 7), n >= 1.
a(n+1) = a(n) + A212152(n)*7^n, n >= 1.
a(n+1) = Sum_{k=0..n} A212152(k)*7^k, n >= 1.
a(n-1)^2*a(n) + 1 == 0 (mod 7^(n-1)), n >= 1 (from 3*a(n)^2* A212152(n) + A210853(n) == 0 (mod 7) and the second-to-last formula from above).
a(n) = 7^n + 1 - A212153(n), n >= 1.

A212153 Approximations up to 7^n for one of the three 7-adic integers (-1)^(1/3).

Original entry on oeis.org

0, 5, 19, 19, 1048, 15454, 82682, 82682, 3376854, 14906456, 135967277, 700917775, 4655571261, 18496858462, 406052900090, 3797168264335, 22787414304107, 188952067152112, 654213095126526, 654213095126526, 57648689021992241, 456610020510052246, 2132247612759904267

Offset: 0

Wolfdieter Lang, May 02 2012

See A210852 for comments and the approximation of one of the other three 7-adic integers (-1)^(1/3), called there u.
The numbers are computed from the recurrence given below in the formula field. This recurrence follows from the formula a(n) = 5^(7^(n-1)) (mod 7^n), n>= 1, which satisfies a(n)^3  + 1 == 0 (mod 7^n), n>=1. a(0) = 0 satisfies this congruence also. The proof can be done by showing that each term in the binomial expansion of (5^(7^(n-1)))^3 + 1 = (2*3^2*7 - 1)^(7^(n-1)) + 1  has a factor 7^n.
a(n) == 5 (mod 7), n>=1. This follows from the formula given above, and 5^(7^(n-1)) == 5 (mod 7), n>=1 (proof by induction).
The digit t(n), n>=0, multiplying 7^n in the 7-adic integer (-1)^(1/3) corresponding to the present sequence is obtained from the (unique) solution of the linear congruence 3*a(n)^2*t(n) + b(n) == 0 (mod 7), n>=1, with b(n):= (a(n)^3 + 1)/7^n = A212154(n). t(0):=5. For these digits see A212155. The 7-adic number is, read from right to left,
...3452150062464440013235234613550254541223240463025 =: v.
  a(n) is obtained from reading v in base 7, and adding the first n terms.
One can show directly that a(n) = 7^n + 1 - x(n), n>=1, with x(n) = A210852(n), and z(n) = 7^n - 1 = 6*A023000(n), n>=0.
Iff a(n+1) = a(n) then t(n) = A212155(n) = 0.
See the Nagell reference given in A210848 for theorems 50 and 52 on p. 87, and formula (6) on page 86, adapted to this case. Because X^3 +1 = 0 (mod 7) has the three simple roots 3, 5 and 6, one has for X(n)^3 +1 == 0 (mod 7^n) exactly three solutions for each n>=1, which can be chosen as x(n) == 3 (mod 7), a(n) == 5 (mod 7) and z(n) == 6 (mod 7) == -1 (mod 7). The x- and z- sequences are given in A210852 and 6*A023000, respectively.
For n>0, a(n) - 1 (== a(n)^2 mod 7^n) and 7^n - a(n) (== a(n)^4 mod 7^n) are the two primitive cubic roots of unity in Z/(7^n Z). - Álvar Ibeas, Feb 20 2017
From Jianing Song, Aug 26 2022: (Start)
a(n) is the solution to x^2 - x + 1 == 0 (mod 7^n) that is congruent to 5 modulo 7 (if n>0).
A210852(n) is the multiplicative inverse of a(n) modulo 7^n. (End)

			a(4) == 19^7 (mod 7^4) = 893871739 (mod 2401) = 1048.
a(4) == 5^343 (mod 7^4) = 1048.
a(4) = 19 + 3*7^3 = 1048.
a(4) = 5*7^0 + 2*7^1 + 0*7^2 + 3*7^3 = 1048.
a(4) = 7^4 + 1 - 1354 = 1048.
a(3) = a(2) = 19 because A212155(2) = 0.

Kenny Lau, Table of n, a(n) for n = 0..1499

Cf. A212155 (digits of (-1)^(1/3)), A210852 (approximations of another cube root of -1), 6*A023000 (approximations of -1).

Cf. A048898, A048899 (approximations of the 5-adic integers sqrt(-1)); A319097, A319098, A319199 (approximations of the 7-adic integers 6^(1/3)).

a:=proc(n) option remember: if n=0 then 0 elif n=1 then 5
else modp(a(n-1)^7, 7^n) fi end proc:

Join[{0}, FoldList[PowerMod[#, 7, 7^#2] &, 5, Range[2, 25]]] (* Paolo Xausa, Jan 14 2025 *)

a(n) = lift((1+sqrt(-3+O(7^n)))/2) \\ Jianing Song, Aug 26 2022

Recurrence: a(n) = a(n-1)^7 (mod 7^n), n>=2, a(0):=0, a(1)=5.
a(n) == 5^(7^(n-1)) (mod 7^n) == 5 (mod 7), n>= 1.
a(n+1) = a(n) + A212155(n)*7^n, n>=1.
a(n+1) = Sum_{k=0..n} A212155(k)*7^k, n>=1.
a(n-1)^2*a(n) + 1 == 0 (mod 7^(n-1)), n>=1 (from 3*a(n)^2*A212155(n) + A212154(n) == 0 (mod 7) and the formula two lines above).
a(n) = 7^n + 1 - A210852(n), n>=1.

A319297 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319097.

Original entry on oeis.org

3, 3, 2, 2, 4, 6, 6, 1, 4, 4, 0, 3, 0, 5, 3, 5, 1, 5, 3, 6, 2, 2, 6, 4, 3, 3, 2, 0, 2, 1, 2, 3, 3, 4, 5, 6, 1, 5, 3, 0, 0, 3, 2, 6, 6, 0, 3, 5, 0, 6, 5, 1, 0, 3, 6, 4, 6, 6, 2, 4, 0, 4, 3, 3, 1, 4, 1, 5, 5, 6, 4, 4, 0, 1, 5, 2, 1, 1, 5, 0, 4, 1, 6, 5, 5, 5, 0, 4

Offset: 0

Jianing Song, Aug 27 2019

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. [Typo corrected by Keyang Li, Nov 04 2024]

			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 = (233)_7, so the first three terms are 3, 3 and 2.

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

Cf. A319097, A319098, A319199.
Digits of p-adic cubic roots:
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A309443 (5-adic, 4^(1/3));
this sequence, A319305, A319555 (7-adic, 6^(1/3));
A321106, A321107, A321108 (13-adic, 5^(1/3)).

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

Equals A319305*(A212152-1) = A319305*A212152^2, where each A-number represents a 7-adic number.

Equals A319555*(A212155-1) = A319555*A212155^2.

A319555 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319199.

Original entry on oeis.org

6, 4, 1, 2, 1, 2, 0, 4, 4, 4, 1, 0, 6, 1, 0, 5, 2, 4, 4, 4, 2, 3, 1, 0, 6, 3, 1, 4, 2, 6, 1, 6, 1, 2, 1, 5, 4, 5, 5, 3, 4, 2, 6, 4, 0, 4, 3, 4, 4, 1, 0, 6, 5, 2, 4, 1, 4, 2, 2, 1, 5, 2, 4, 4, 2, 5, 4, 6, 5, 1, 0, 1, 6, 1, 1, 4, 0, 6, 3, 4, 4, 2, 3, 4, 0, 0, 4, 4

Offset: 0

Jianing Song, Aug 27 2019

For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 13). If k is a cube in 7-adic field, then k has exactly three cubic roots.

			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 = (146)_7, so the first three terms are 6, 4 and 1.

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

Cf. A319097, A319098, A319199.
Digits of p-adic cubic roots:
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A309443 (5-adic, 4^(1/3));
A319297, A319305, this sequence (7-adic, 6^(1/3));
A321106, A321107, A321108 (13-adic, 5^(1/3)).

PARI
```
a(n) = lift(sqrtn(6+O(7^(n+1)), 3))\7^n
```

Equals A319297*(A212152-1) = A319297*A212152^2, where each A-number represents a 7-adic number.

Equals A319305*(A212155-1) = A319305*A212155^2.

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).

Original entry on oeis.org

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

Offset: 0

Jianing Song, Aug 27 2019

nonn

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.

			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.

Wikipedia, p-adic number

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)).

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

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.

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).

Original entry on oeis.org

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

Offset: 0

Jianing Song, Aug 27 2019

nonn

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.

			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.

Wikipedia, p-adic number

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)).

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

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.

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

Jianing Song, Aug 30 2019

nonn

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

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

Wikipedia, p-adic number

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)).

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

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

Jianing Song, Aug 30 2019

nonn

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

			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.

Wikipedia, p-adic number

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)).

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

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

Jianing Song, Aug 30 2019

nonn

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

			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.

Wikipedia, p-adic number

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)).

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

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).

cp's OEIS Frontend

A319305 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319098.

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A210852 Approximations up to 7^n for one of the three 7-adic integers (-1)^(1/3).

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A212153 Approximations up to 7^n for one of the three 7-adic integers (-1)^(1/3).

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A319297 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319097.

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A319555 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319199.

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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).

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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).

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