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A332124 a(n) = 2*(10^(2n+1)-1)/9 + 2*10^n. 3

%I #11 Mar 06 2023 08:17:11

%S 4,242,22422,2224222,222242222,22222422222,2222224222222,

%T 222222242222222,22222222422222222,2222222224222222222,

%U 222222222242222222222,22222222222422222222222,2222222222224222222222222,222222222222242222222222222,22222222222222422222222222222,2222222222222224222222222222222

%N a(n) = 2*(10^(2n+1)-1)/9 + 2*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) = 2*A138148(n) + 4*10^n = A002276(2n+1) + 2*10^n = 2*A332112(n).

%F G.f.: (4 - 202*x)/((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 A332124 := n -> 2*((10^(2*n+1)-1)/9+10^n);

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

%t Table[FromDigits[Join[PadRight[{},n,2],{4},PadRight[{},n,2]]],{n,0,20}] (* or *) LinearRecurrence[{111,-1110,1000},{4,242,22422},20](* _Harvey P. Dale_, Mar 06 2023 *)

%o (PARI) apply( {A332124(n)=(10^(n*2+1)\9+10^n)*2}, [0..15])

%o (Python) def A332124(n): return (10**(n*2+1)//9+10**n)*2

%Y Cf. A002275 (repunits R_n = (10^n-1)/9), A002276 (2*R_n), A011557 (10^n).

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

%Y Cf. A332114 .. A332194 (variants with different repeated digit 1, ..., 9).

%Y Cf. A332120 .. A332129 (variants with different middle digit 0, ..., 9).

%K nonn,base,easy

%O 0,1

%A _M. F. Hasler_, Feb 09 2020

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Last modified March 29 00:26 EDT 2024. Contains 371264 sequences. (Running on oeis4.)