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%I #94 Mar 14 2023 09:31:00
%S 0,1,8,57,400,2801,19608,137257,960800,6725601,47079208,329554457,
%T 2306881200,16148168401,113037178808,791260251657,5538821761600,
%U 38771752331201,271402266318408,1899815864228857,13298711049602000
%N a(n) = (7^n - 1)/6.
%C Apart from a(0), numbers of the form 11...11 (i.e., repunits) in base 7.
%C 7^(floor(7^n/6)) is the highest power of 7 dividing (7^n)!. - _Benoit Cloitre_, Feb 04 2002
%C 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
%C 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
%C 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
%C From _Wolfdieter Lang_, May 02 2012: (Start)
%C 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.
%C 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.
%C (End)
%H Vincenzo Librandi, <a href="/A023000/b023000.txt">Table of n, a(n) for n = 0..1000</a>
%H Roger B. Eggleton, <a href="http://dx.doi.org/10.1155/2015/216475">Maximal Midpoint-Free Subsets of Integers</a>, International Journal of Combinatorics Volume 2015, Article ID 216475, 14 pages.
%H Wolfdieter Lang, <a href="/A023000/a023000.pdf">Notes on certain inhomogeneous three term recurrences</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Repunit.html">Repunit</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/As_I_was_going_to_St_Ives">As I was going to St Ives</a>.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-7).
%F From _R. J. Mathar_, Jun 21 2009: (Start)
%F a(n) = 8*a(n-1) - 7*a(n-2).
%F G.f.: x/((1-x)*(1-7*x)). (End)
%F From _Wolfdieter Lang_, Oct 18 2010: (Start)
%F a(n) = 6*a(n-1) + 7*a(n-2) + 2, a(0)=0, a(1)=1.
%F 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)
%F a(n) = 7*a(n-1) + 1 (with a(0)=0). - _Vincenzo Librandi_, Nov 19 2010
%F a(n) = a(n-1) + 7^(n-1), with a(0)=0, n >= 1. - See a _Wolfdieter Lang_ comment above, May 02 2012
%F E.g.f.: exp(4*x)*sinh(3*x)/3. - _Stefano Spezia_, Mar 11 2023
%t LinearRecurrence[{8, -7}, {0, 1}, 30] (* _Vincenzo Librandi_, Nov 08 2012 *)
%t (7^Range[0,20]-1)/6 (* _Harvey P. Dale_, Aug 03 2020 *)
%o (Sage)
%o def a(n): return (7**n-1)//6
%o [a(n) for n in range(66)] # show terms
%o # _Joerg Arndt_, May 28 2012
%o (PARI) a(n)=(7^n-1)/6; /* _Joerg Arndt_, May 28 2012 */
%o (Maxima) A023000(n):=floor((7^n-1)/6)$ makelist(A023000(n),n,0,30); /* _Martin Ettl_, Nov 05 2012 */
%o (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
%Y Cf. A210852, A212153, A212156.
%K easy,nonn
%O 0,3
%A _David W. Wilson_
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