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a(1)=1, a(n+1) = 10*a(n)+2*n-1
1

%I #16 Mar 04 2024 01:15:11

%S 1,13,135,1357,13579,135801,1358023,13580245,135802467,1358024689,

%T 13580246911,135802469133,1358024691355,13580246913577,

%U 135802469135799,1358024691358021,13580246913580243,135802469135802465,1358024691358024687,13580246913580246909

%N a(1)=1, a(n+1) = 10*a(n)+2*n-1

%C Third column of A164851. The repeating pattern corresponds to the decimal expansion of 11/81 = 0.13580246913580246...

%H Colin Barker, <a href="/A179619/b179619.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 (12,-21,10).

%F a(1)=1, a(2)=13, a(3)=135, a(n) = 12*a(n-1)-21*a(n-2)+10*a(n-3). - _Harvey P. Dale_, Aug 16 2012

%F G.f.: -x*(x+1) / ((x-1)^2*(10*x-1)). - _Colin Barker_, Oct 03 2015

%t RecurrenceTable[{a[1] == 1, a[n] == 10 a[n - 1] + 2 n - 1}, a, {n,

%t 30}] (* or *) LinearRecurrence[{12,-21,10},{1,13,135},30] (* _Harvey P. Dale_, Aug 16 2012 *)

%o (PARI) Vec(-x*(x+1)/((x-1)^2*(10*x-1)) + O(x^30)) \\ _Colin Barker_, Oct 03 2015

%Y Cf. A164851.

%K nonn,easy

%O 1,2

%A _Mark Dols_, Jul 20 2010