A212156 -id:A212156 - cp's OEIS frontend

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

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

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

Original entry on oeis.org

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.

cp's OEIS Frontend

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

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