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

Author

David W. Wilson

Keywords

• easy
• nonn

Comments

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)

Links

• 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).

Crossrefs

Cf. A210852, A212153, A212156.

Programs

Magma

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

Mathematica

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

Maxima

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

PARI

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

Sage

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

Formula

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