|
%I #25 Apr 23 2022 14:55:42
%S 1,11,102,922,8303,74733,672604,6053444,54481005,490329055,4412961506,
%T 39716653566,357449882107,3217048938977,28953440450808,
%U 260580964057288,2345228676515609,21107058088640499
%N a(1)=1; for n>1, a(n) = 9*a(n-1)+n.
%H Seiichi Manyama, <a href="/A014832/b014832.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (11,-19,9).
%F a(n) = (9^(n+1)-8*n-9)/64. - _Rolf Pleisch_, Oct 22 2010
%F a(1)=1, a(2)=11, a(3)=102; for n>3, a(n) = 11*a(n-1) - 19*a(n-2) + 9*a(n-3). - _Harvey P. Dale_, May 01 2012
%F G.f.: -(x/((x-1)^2*(9*x-1))). - _Harvey P. Dale_, May 01 2012
%F a(n) = Sum_{i=0..n-1} 8^i*binomial(n+1,n-1-i). [_Bruno Berselli_, Nov 13 2015]
%e For n=5, a(5) = 1*15 + 8*20 + 8^2*15 + 8^3*6 + 8^4*1 = 8303. [_Bruno Berselli_, Nov 13 2015]
%p a:=n->sum((9^(n-j)-1)/8,j=0..n): seq(a(n), n=1..18); # _Zerinvary Lajos_, Jan 15 2007
%p a:= n-> (Matrix([[1,0,1],[1,1,1],[0,0,9]])^n)[2,3]: seq(a(n), n=1..18); # _Alois P. Heinz_, Aug 06 2008
%t RecurrenceTable[{a[1]==1,a[n]==9a[n-1]+n},a,{n,20}] (* or *) LinearRecurrence[ {11,-19,9},{1,11,102},20] (* _Harvey P. Dale_, May 01 2012 *)
%Y Cf. A001018, A104712.
%K nonn,easy
%O 1,2
%A _N. J. A. Sloane_
|
