A014832 a(1)=1; for n>1, a(n) = 9*a(n-1) + n.

1, 11, 102, 922, 8303, 74733, 672604, 6053444, 54481005, 490329055, 4412961506, 39716653566, 357449882107, 3217048938977, 28953440450808, 260580964057288, 2345228676515609, 21107058088640499, 189963522797764510, 1709671705179880610, 15387045346618925511, 138483408119570329621

Offset

1,2

Links

Seiichi Manyama, Table of n, a(n) for n = 1..1000

Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See p. 18.

Index entries for linear recurrences with constant coefficients, signature (11,-19,9).

Formula

a(n) = (9^(n+1) - 8*n - 9)/64. - Rolf Pleisch, Oct 22 2010

From Harvey P. Dale, May 01 2012: (Start)

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).

G.f.: -x/((x-1)^2*(9*x-1)). (End)

a(n) = Sum_{i=0..n-1} 8^i*binomial(n+1,n-1-i). - Bruno Berselli, Nov 13 2015

E.g.f.: exp(x)*(9*exp(8*x) - 8*x - 9)/64. - Elmo R. Oliveira, Mar 29 2025

Example

For n=5, a(5) = 1*15 + 8*20 + 8^2*15 + 8^3*6 + 8^4*1 = 8303. [Bruno Berselli, Nov 13 2015]

Maple

a:=n->sum((9^(n-j)-1)/8, j=0..n): seq(a(n), n=1..18); # Zerinvary Lajos, Jan 15 2007
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

Mathematica

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 *)

Crossrefs

Cf. A001018, A104712.

Sequence in context: A037700 A037609 A055150 * A048441 A099294 A081552

Adjacent sequences: A014829 A014830 A014831 * A014833 A014834 A014835

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved