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a(n) = 4*(10^(2*n+1)-1)/9 - 3*10^n.
2

%I #9 Aug 17 2020 14:27:38

%S 1,414,44144,4441444,444414444,44444144444,4444441444444,

%T 444444414444444,44444444144444444,4444444441444444444,

%U 444444444414444444444,44444444444144444444444,4444444444441444444444444,444444444444414444444444444,44444444444444144444444444444,4444444444444441444444444444444

%N a(n) = 4*(10^(2*n+1)-1)/9 - 3*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) = 4*A138148(n) + 1*10^n = A002278(2n+1) - 3*10^n.

%F G.f.: (1 + 303*x - 700*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 A332141 := n -> 4*(10^(2*n+1)-1)/9-3*10^n;

%t Array[4 (10^(2 # + 1)-1)/9 - 3*10^# &, 15, 0]

%t LinearRecurrence[{111,-1110,1000},{1,414,44144},20] (* or *) Table[ FromDigits[Join[PadRight[{},n,4],{1},PadRight[{},n,4]]],{n,0,20}](* _Harvey P. Dale_, Aug 17 2020 *)

%o (PARI) apply( {A332141(n)=10^(n*2+1)\9*4-3*10^n}, [0..15])

%o (Python) def A332141(n): return 10**(n*2+1)//9*4-3*10**n

%Y Cf. A002275 (repunits R_n = (10^n-1)/9), A002278 (4*R_n), A011557 (10^n).

%Y Cf. A138148 (cyclops numbers with binary digits), A002113 (palindromes).

%Y Cf. A332121 .. A332191 (variants with different repeated digit 2, ..., 9).

%Y Cf. A332140 .. A332149 (variants with different middle digit 0, ..., 9).

%K nonn,base,easy

%O 0,2

%A _M. F. Hasler_, Feb 09 2020