login
a(n) = (10^n + 53)/9
1

%I #20 Jul 14 2015 09:43:02

%S 7,17,117,1117,11117,111117,1111117,11111117,111111117,1111111117,

%T 11111111117,111111111117,1111111111117,11111111111117,

%U 111111111111117,1111111111111117,11111111111111117,111111111111111117

%N a(n) = (10^n + 53)/9

%C a(n) are also n-1 units followed by a seven.

%H Markus Tervooren, <a href="http://factordb.com/search.php?query=%2810%5Ex%2B53%29%2F9">Factorizations of (1)w7</a>

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

%F a(n) = 9*a(n-1)-8*a(n-2)+2*10^(n-2) for n>2, a(1)=7, a(2)=17. [_Vincenzo Librandi_, Aug 02 2010]

%F a(n) = 11*a(n-1)-10*a(n-2). G.f.: -x*(60*x-7)/((x-1)*(10*x-1)). [_Colin Barker_, Jan 24 2013]

%t Table[FromDigits[PadLeft[{7}, n, 1]], {n, 20}] (* _Harvey P. Dale_, Oct 17 2011 *)

%K nonn,easy

%O 1,1

%A _Ivan Panchenko_, Sep 10 2009