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a(n) = (2*10^n+61)/9
0

%I #14 Apr 09 2019 15:35:20

%S 9,29,229,2229,22229,222229,2222229,22222229,222222229,2222222229,

%T 22222222229,222222222229,2222222222229,22222222222229,

%U 222222222222229,2222222222222229,22222222222222229,222222222222222229

%N a(n) = (2*10^n+61)/9

%C n-1 2's followed by a 9.

%H Markus Tervooren, <a href="http://factordb.com/search.php?query=%282%2A10%5Ex%2B61%29%2F9">Factorizations of (2)w9</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11, -10).

%F a(n)= 11*a(n-1) -10*a(n-2). G.f.: x*(-9+70*x)/((10*x-1)*(1-x)). [From _R. J. Mathar_, Feb 21 2010]

%F a(n)=10*a(n-1)-61, (with a(1)=9) [From _Vincenzo Librandi_, Feb 24 2010]

%t Table[10FromDigits[PadRight[{},n,2]]+9,{n,0,20}] (* or *) LinearRecurrence[ {11,-10},{9,29},20] (* _Harvey P. Dale_, Apr 09 2019 *)

%K easy,nonn

%O 1,1

%A _Ivan Panchenko_, Feb 14 2010

%E Extended by _R. J. Mathar_, Feb 21 2010