%I #7 May 31 2023 15:16:11
%S 9,595,55955,5559555,555595555,55555955555,5555559555555,
%T 555555595555555,55555555955555555,5555555559555555555,
%U 555555555595555555555,55555555555955555555555,5555555555559555555555555,555555555555595555555555555,55555555555555955555555555555,5555555555555559555555555555555
%N a(n) = 5*(10^(2*n+1)-1)/9 + 4*10^n.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (111,-1110,1000).
%F a(n) = 5*A138148(n) + 9*10^n = A002279(2n+1) + 4*10^n.
%F G.f.: (9 - 404*x - 100*x^2)/((1 - x)(1 - 10*x)(1 - 100*x)).
%F a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.
%p A332159 := n -> 5*(10^(2*n+1)-1)/9+4*10^n;
%t Array[5 (10^(2 # + 1)-1)/9 + 4*10^# &, 15, 0]
%t Table[FromDigits[Join[PadRight[{},n,5],PadRight[{9},n+1,5]]],{n,0,20}] (* or *) LinearRecurrence[ {111,-1110,1000},{9,595,55955},20] (* _Harvey P. Dale_, May 31 2023 *)
%o (PARI) apply( {A332159(n)=10^(n*2+1)\9*5+4*10^n}, [0..15])
%o (Python) def A332159(n): return 10**(n*2+1)//9*5+4*10**n
%Y Cf. A002275 (repunits R_n = (10^n-1)/9), A002279 (5*R_n), A011557 (10^n).
%Y Cf. A138148 (cyclops numbers with binary digits), A002113 (palindromes).
%Y Cf. A332119 .. A332189 (variants with different repeated digit 1, ..., 8).
%Y Cf. A332150 .. A332159 (variants with different middle digit 0, ..., 9).
%K nonn,base,easy
%O 0,1
%A _M. F. Hasler_, Feb 09 2020