|
|
A091030
|
|
Partial sums of powers of 13 (A001022).
|
|
44
|
|
|
1, 14, 183, 2380, 30941, 402234, 5229043, 67977560, 883708281, 11488207654, 149346699503, 1941507093540, 25239592216021, 328114698808274, 4265491084507563, 55451384098598320, 720867993281778161
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
13^a(n) is highest power of 13 dividing (13^n)!.
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
|
|
|
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) = Sum_{k=0...n-1} 12^k*binomial(n,n-1-k). - Bruno Berselli, Nov 12 2015
|
|
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
|
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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|