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

%I #7 Feb 11 2020 08:09:08

%S 5,454,44544,4445444,444454444,44444544444,4444445444444,

%T 444444454444444,44444444544444444,4444444445444444444,

%U 444444444454444444444,44444444444544444444444,4444444444445444444444444,444444444444454444444444444,44444444444444544444444444444,4444444444444445444444444444444

%N a(n) = 4*(10^(2*n+1)-1)/9 + 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) + 5*10^n = A002278(2n+1) + 10^n.

%F G.f.: (5 - 101*x - 300*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 A332145 := n -> 4*(10^(2*n+1)-1)/9+10^n;

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

%o (PARI) apply( {A332145(n)=10^(n*2+1)\9*4+10^n}, [0..15])

%o (Python) def A332145(n): return 10**(n*2+1)//9*4+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. A332115 .. A332195 (variants with different repeated digit 1, ..., 9).

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

%K nonn,base,easy

%O 0,1

%A _M. F. Hasler_, Feb 09 2020