A023000 - OEIS

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

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)

Eric Weisstein's World of Mathematics, Repunit.

a(n) = 8*a(n-1) - 7*a(n-2).

G.f.: x/((1-x)*(1-7*x)). (End)

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) = 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

LinearRecurrence[{8, -7}, {0, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)

(Sage)

def a(n): return (7**n-1)//6

[a(n) for n in range(66)] # show terms

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

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

(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