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a(n) = 5*(10^(2*n+1)-1)/9 - 3*10^n.
1

%I #6 Feb 11 2020 08:15:07

%S 2,525,55255,5552555,555525555,55555255555,5555552555555,

%T 555555525555555,55555555255555555,5555555552555555555,

%U 555555555525555555555,55555555555255555555555,5555555555552555555555555,555555555555525555555555555,55555555555555255555555555555,5555555555555552555555555555555

%N a(n) = 5*(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) = 5*A138148(n) + 2*10^n = A002279(2n+1) - 3*10^n.

%F G.f.: (2 + 303*x - 800*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 A332152 := n -> 5*(10^(2*n+1)-1)/9-3*10^n;

%t Array[5 (10^(2 # + 1)-1)/9 - 3*10^# &, 15, 0]

%o (PARI) apply( {A332152(n)=10^(n*2+1)\9*5-3*10^n}, [0..15])

%o (Python) def A332152(n): return 10**(n*2+1)//9*5-3*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. A332112 .. A332192 (variants with different repeated digit 1, ..., 9).

%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