OFFSET
0,2
COMMENTS
11^a(n) is highest power of 11 dividing (11^(n+1))!.
Partial sums of powers of 11 (A001020).
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=11, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=det(A). - Milan Janjic, Feb 21 2010
Let A be the Hessenberg matrix of the order n, defined by: A[1,j]=1, A[i,i]:=12, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=2, a(n-1)=(-1)^n*charpoly(A,1). - Milan Janjic, Feb 21 2010
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..950
Eric Weisstein's World of Mathematics, Repunit
Index entries for linear recurrences with constant coefficients, signature (12,-11).
FORMULA
a(n) = Sum_{k=0..n} 11^k = (11^(n+1) - 1)/10.
G.f.: (1/(1-11*x) - 1/(1-x))/(10*x) = 1/((1-11*x)*(1-x)).
a(0)=1, a(n) = 11*a(n-1) + 1. - Vincenzo Librandi, Feb 05 2011
a(0)=0, a(1)=1, a(n) = 12*a(n-1) - 11*a(n-2). - Harvey P. Dale, Apr 05 2012
E.g.f.: exp(x)*(11*exp(10*x) - 1)/10. - Stefano Spezia, Mar 11 2023
MATHEMATICA
(11^Range[0, 20]-1)/10 (* or *) LinearRecurrence[{12, -11}, {0, 1}, 20] (* Harvey P. Dale, Apr 05 2012 *)
PROG
(Sage) [lucas_number1(n, 12, 11) for n in range(1, 19)] # Zerinvary Lajos, Apr 27 2009
(Sage) [gaussian_binomial(n, 1, 11) for n in range(1, 19)] # Zerinvary Lajos, May 28 2009
(Maxima) A016123(n):=(11^(n+1)-1)/10$
makelist(A016123(n), n, 0, 30); /* Martin Ettl, Nov 05 2012 */
(PARI) a(n)=(11^(n+1)-1)/10 \\ Charles R Greathouse IV, Sep 24 2015
(Magma) [(11^(n+1)-1)/10: n in [0..30]]; // G. C. Greubel, Feb 21 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
Title edited by Daniel Forgues, Jul 08 2011
STATUS
approved