A212153 -id:A212153 - cp's OEIS frontend

A212154 a(n) = (A212153(n)^3 + 1)/7^n, n >= 0.

1, 18, 140, 20, 479393, 219600095, 4804461081, 686351583, 6679631931865, 82080661415031, 8898622841908566, 174149720118385232, 7290250572352382182, 65315972853762054047, 98713213404986046050649, 11532114009920222592500432, 356054521382275298405890644, 28999349909865958163356878647

Offset: 0

Wolfdieter Lang, May 02 2012

a(n) is an integer because A212153(n) is one of the three solutions of X(n)^3+1 == 0 (mod 7^n), namely the one satisfying also X(n) == 5 (mod 7).

See the comments on A210853, and the Nagell reference given in A210848.

			a(0) = 1/1 = 1.
a(3) = (19^3 + 1)/7^3 = 6860/343  = 20, (b(3) = 19^7 (mod 7^3) = 19).

Paolo Xausa, Table of n, a(n) for n = 0..500

Cf. A210848, A210849 (the p=5 case), A210853, A212153, A212156.

Join[{1}, MapIndexed[(#^3 + 1)/7^#2[[1]] &, FoldList[PowerMod[#, 7, 7^#2] &, 5, Range[2, 20]]]] (* Paolo Xausa, Jan 14 2025 *)

a(n) = (A212153(n)^3 + 1)/7^n.

A023000 a(n) = (7^n - 1)/6.

Original entry on oeis.org

0, 1, 8, 57, 400, 2801, 19608, 137257, 960800, 6725601, 47079208, 329554457, 2306881200, 16148168401, 113037178808, 791260251657, 5538821761600, 38771752331201, 271402266318408, 1899815864228857, 13298711049602000

Offset: 0

David W. Wilson

Apart from a(0), numbers of the form 11...11 (i.e., repunits) in base 7.
7^(floor(7^n/6)) is the highest power of 7 dividing (7^n)!. - Benoit Cloitre, Feb 04 2002
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=7, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=8, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^n*charpoly(A,1). - Milan Janjic, Feb 21 2010
This is the sequence A(0,1;6,7;2) = A(0,1;8,-7;0) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
From Wolfdieter Lang, May 02 2012: (Start)
6*a(n) =: z(n) gives the approximation up to 7^n for one of the three 7-adic integers (-1)^(1/3), i.e. z(n)^3 + 1 == 0 (mod 7^n), n>=0, and z(n) == 6 (mod 7) == -1 (mod 7), n>=1. The companion sequences are x(n) = A210852(n) and y(n) = A212153(n). This leads to a(n) == 1 (mod 7) for n>=1 (this is also clear from some of the formulas given below). Also 216*a(n)^3 + 1 == 0 (mod 7^n), n>=0, as well as 3*216*a(n)^2 + A212156(n) == 0 (mod 7^n), n>=0. a(n) = 6^(7^(n-1)-1) (mod 7^n), n>=1. A recurrence is a(n) = a(n-1) + 7^(n-1), with a(0)=0, for n>=1.
Also a(n) = (1/6)*(6*a(n-1))^7 (mod 7^n) with a(1)=1 for n>=1. Finally, 6^3*a(n-1)*a(n)^2 + 1 == 0 (mod 7^(n-1)), n>=1.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Shaoshi Chen, Hanqian Fang, Sergey Kitaev, and Candice X.T. Zhang, Patterns in Multi-dimensional Permutations, arXiv:2411.02897 [math.CO], 2024. See pp. 2, 17.
Carlos M. da Fonseca and Anthony G. Shannon, A formal operator involving Fermatian numbers, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
Roger B. Eggleton, Maximal Midpoint-Free Subsets of Integers, International Journal of Combinatorics Volume 2015, Article ID 216475, 14 pages.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
D. C. Santos, E. A. Costa, and P. M. M. C. Catarino, On Gersenne Sequence: A Study of One Family in the Horadam-Type Sequence, Axioms 14, 203, (2025). See p. 4.
Eric Weisstein's World of Mathematics, Repunit.
Wikipedia, As I was going to St Ives.
Index entries for linear recurrences with constant coefficients, signature (8,-7).

Cf. A210852, A212153, A212156.

[n le 2 select n-1 else 8*Self(n-1) - 7*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012

LinearRecurrence[{8, -7}, {0, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
(7^Range[0,20]-1)/6 (* Harvey P. Dale, Aug 03 2020 *)

A023000(n):=floor((7^n-1)/6)$  makelist(A023000(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

a(n)=(7^n-1)/6;  /* Joerg Arndt, May 28 2012 */

def a(n): return (7**n-1)//6
[a(n) for n in range(66)] # show terms
# Joerg Arndt, May 28 2012

From R. J. Mathar, Jun 21 2009: (Start)
a(n) = 8*a(n-1) - 7*a(n-2).
G.f.: x/((1-x)*(1-7*x)). (End)
From Wolfdieter Lang, Oct 18 2010: (Start)
a(n) = 6*a(n-1) + 7*a(n-2) + 2, a(0)=0, a(1)=1.
a(n) = 7*a(n-1) + a(n-2) - 7*a(n-3) = 9*a(n-1) - 15*a(n-2) + 7*a(n-3), a(0)=0, a(1)=1, a(2)=8. Observation by G. Detlefs. See the W. Lang comment and link. (End)
a(n) = 7*a(n-1) + 1 (with a(0)=0). - Vincenzo Librandi, Nov 19 2010
a(n) = a(n-1) + 7^(n-1), with a(0)=0, n >= 1. - See a Wolfdieter Lang comment above, May 02 2012
E.g.f.: exp(4*x)*sinh(3*x)/3. - Stefano Spezia, Mar 11 2023

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.

A309450 The successive approximations up to 7^n for 7-adic integer 2^(1/5).

Original entry on oeis.org

0, 4, 46, 95, 1124, 15530, 82758, 435705, 4553420, 27612624, 269734266, 1682110511, 9591417483, 9591417483, 9591417483, 4078929854577, 23069175894349, 122767967603152, 1053290023551980, 9195358013104225, 77588729125343083, 237173261720567085, 1354264989887135099

Offset: 0

Seiichi Manyama, Aug 03 2019

nonn

			a(1) = (   4)_7 = 4,
a(2) = (  64)_7 = 46,
a(3) = ( 164)_7 = 95,
a(4) = (3164)_7 = 1124.

Robert Israel, Table of n, a(n) for n = 0..1182

Cf. A309445.
Expansions of p-adic integers:
A290567 (5-adic, 2^(1/3));
A290800, A290802 (7-adic, sqrt(-6));
A290806, A290809 (7-adic, sqrt(-5));
A290803, A290804 (7-adic, sqrt(-3));
A210852, A212153 (7-adic, (1+sqrt(-3))/2);
A290557, A290559 (7-adic, sqrt(2));
A309451 (7-adic, 3^(1/5));
A309452 (7-adic, 4^(1/5));
A309453 (7-adic, 5^(1/5));
A309454 (7-adic, 6^(1/5)).

A:= op([1,3],padic:-rootp(x^5 -2,  7, 25)):
seq(add(A[i]*10^(i-1),i=1..n),n=0..25); # Robert Israel, Aug 04 2019

PARI
```
{a(n) = truncate((2+O(7^n))^(1/5))}
```

a(0) = 0 and a(1) = 4, a(n) = a(n-1) + (a(n-1)^5 - 2) mod 7^n for n > 1.

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.

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).

Original entry on oeis.org

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

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 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.

			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.

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, this sequence, 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) = 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.

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.

A309451 The successive approximations up to 7^n for 7-adic integer 3^(1/5).

Original entry on oeis.org

0, 5, 26, 75, 1104, 3505, 20312, 20312, 4961570, 28020774, 229788809, 512264058, 2489590801, 71696026806, 71696026806, 71696026806, 19061942066578, 218459525484184, 451090039471391

Offset: 0

Seiichi Manyama, Aug 03 2019

nonn

			a(1) = (   5)_7 = 5,
a(2) = (  35)_7 = 26,
a(3) = ( 135)_7 = 75,
a(4) = (3135)_7 = 1104.

Cf. A309446.
Expansions of p-adic integers:
A290568 (5-adic, 3^(1/3));
A290800, A290802 (7-adic, sqrt(-6));
A290806, A290809 (7-adic, sqrt(-5));
A290803, A290804 (7-adic, sqrt(-3));
A210852, A212153 (7-adic, (1+sqrt(-3))/2);
A290557, A290559 (7-adic, sqrt(2));
A309450 (7-adic, 2^(1/5));
A309452 (7-adic, 4^(1/5));
A309453 (7-adic, 5^(1/5));
A309454 (7-adic, 6^(1/5)).

PARI
```
{a(n) = truncate((3+O(7^n))^(1/5))}
```

a(0) = 0 and a(1) = 5, a(n) = a(n-1) + 2 * (a(n-1)^5 - 3) mod 7^n for n > 1.

A309452 The successive approximations up to 7^n for 7-adic integer 4^(1/5).

Original entry on oeis.org

0, 2, 9, 107, 450, 450, 67678, 655923, 2303009, 13832611, 54186218, 1749037712, 13612998170, 27454285371, 124343295778, 4193681732872, 18436366262701, 217833949680307, 1380986519616342, 3009400117526791, 3009400117526791, 162593932712750793, 3513869117212454835

Offset: 0

Seiichi Manyama, Aug 03 2019

nonn

			a(1) = (   2)_7 = 2,
a(2) = (  12)_7 = 9,
a(3) = ( 212)_7 = 107,
a(4) = (1212)_7 = 450.

Cf. A309447.
Expansions of p-adic integers:
A290800, A290802 (7-adic, sqrt(-6));
A290806, A290809 (7-adic, sqrt(-5));
A290803, A290804 (7-adic, sqrt(-3));
A210852, A212153 (7-adic, (1+sqrt(-3))/2);
A290557, A290559 (7-adic, sqrt(2));
A309450 (7-adic, 2^(1/5));
A309451 (7-adic, 3^(1/5));
A309453 (7-adic, 5^(1/5));
A309454 (7-adic, 6^(1/5)).

PARI
```
{a(n) = truncate((4+O(7^n))^(1/5))}
```

a(0) = 0 and a(1) = 2, a(n) = a(n-1) + 2 * (a(n-1)^5 - 4) mod 7^n for n > 1.

A309453 The successive approximations up to 7^n for 7-adic integer 5^(1/5).

Original entry on oeis.org

0, 3, 45, 339, 1368, 8571, 42185, 630430, 4748145, 27807349, 27807349, 1722658843, 13586619301, 41269193703, 235047214517, 2269716433064, 30755085492722, 230152668910328, 928044210871949, 2556457808782398, 36753143364901827, 196337675960125829, 2430521132293261857

Offset: 0

Seiichi Manyama, Aug 03 2019

nonn

			a(1) = (   3)_7 = 3,
a(2) = (  63)_7 = 45,
a(3) = ( 663)_7 = 339,
a(4) = (3663)_7 = 1368.

Cf. A309448.
Expansions of p-adic integers:
A290800, A290802 (7-adic, sqrt(-6));
A290806, A290809 (7-adic, sqrt(-5));
A290803, A290804 (7-adic, sqrt(-3));
A210852, A212153 (7-adic, (1+sqrt(-3))/2);
A290557, A290559 (7-adic, sqrt(2));
A309450 (7-adic, 2^(1/5));
A309451 (7-adic, 3^(1/5));
A309452 (7-adic, 4^(1/5));
A309454 (7-adic, 6^(1/5)).

PARI
```
{a(n) = truncate((5+O(7^n))^(1/5))}
```

a(0) = 0 and a(1) = 3, a(n) = a(n-1) + (a(n-1)^5 - 5) mod 7^n for n > 1.

cp's OEIS Frontend

A212154 a(n) = (A212153(n)^3 + 1)/7^n, n >= 0.

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A023000 a(n) = (7^n - 1)/6.

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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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A309450 The successive approximations up to 7^n for 7-adic integer 2^(1/5).

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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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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).

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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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