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a(n) = floor(10^(n+1)/81).
11

%I #20 Jan 29 2025 07:10:41

%S 0,1,12,123,1234,12345,123456,1234567,12345679,123456790,1234567901,

%T 12345679012,123456790123,1234567901234,12345679012345,

%U 123456790123456,1234567901234567,12345679012345679,123456790123456790,1234567901234567901,12345679012345679012,123456790123456790123

%N a(n) = floor(10^(n+1)/81).

%H Vincenzo Librandi, <a href="/A057932/b057932.txt">Table of n, a(n) for n = 0..300</a>

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (10,0,0,0,0,0,0,0,1,-10).

%t Floor[10^(Range[0, 25] + 1)/81] (* _Paolo Xausa_, Jan 29 2025 *)

%o (Magma) [Floor(10^(n+1)/81): n in [1..20]]; // _Vincenzo Librandi_, Jul 07 2011

%o (PARI) a(n)=10^(n+1)\81 \\ _Charles R Greathouse IV_, Jul 07 2011

%Y Cf. A014824, A057933.

%K nonn,easy,changed

%O 0,3

%A _Henry Bottomley_, Oct 04 2000