OFFSET
1,2
COMMENTS
13^a(n) is highest power of 13 dividing (13^n)!.
For analogs with primes 2, 3, 5, 7 and 11 see A000225, A003462, A003463, A023000 and A016123 respectively.
Let A be the Hessenberg matrix of the order n, defined by: A[1,j]=1,A[i,i]:=13, (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]:=14, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=(-1)^(n)*charpoly(A,1). - Milan Janjic, Feb 21 2010
LINKS
Delbert L. Johnson, Table of n, a(n) for n = 1..898
Index entries for linear recurrences with constant coefficients, signature (14,-13).
FORMULA
G.f.: x/((1-13*x)*(1-x)) = (1/(1-13*x) - 1/(1-x))/12.
a(n) = Sum_{k=0..n-1} 13^k = (13^n-1)/12.
a(n) = 13*a(n-1)+1 for n>1, a(1)=1. - Vincenzo Librandi, Feb 05 2011
a(n) = Sum_{k=0...n-1} 12^k*binomial(n,n-1-k). - Bruno Berselli, Nov 12 2015
E.g.f.: exp(x)*(exp(12*x) - 1)/12. - Stefano Spezia, Mar 11 2023
EXAMPLE
For n=6, a(6) = 1*6 + 12*15 + 144*20 + 1728*15 + 20736*6 + 248832*1 = 402234. - Bruno Berselli, Nov 12 2015
MAPLE
a:=n->sum(13^(n-j), j=1..n): seq(a(n), n=1..17); # Zerinvary Lajos, Jan 04 2007
MATHEMATICA
Table[13^n, {n, 0, 16}] // Accumulate (* Jean-François Alcover, Jul 05 2013 *)
LinearRecurrence[{14, -13}, {1, 14}, 20] (* Harvey P. Dale, Jan 19 2024 *)
PROG
(Sage) [gaussian_binomial(n, 1, 13) for n in range(1, 18)] # Zerinvary Lajos, May 28 2009
(Sage) [(13^n-1)/12 for n in (1..30)] # Bruno Berselli, Nov 12 2015
(PARI) a(n)=([0, 1; -13, 14]^(n-1)*[1; 14])[1, 1] \\ Charles R Greathouse IV, Sep 24 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jan 23 2004
STATUS
approved