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a(n) = 5*10^n - 1.
8

%I #42 Sep 17 2024 15:41:43

%S 4,49,499,4999,49999,499999,4999999,49999999,499999999,4999999999,

%T 49999999999,499999999999,4999999999999,49999999999999,

%U 499999999999999,4999999999999999,49999999999999999,499999999999999999,4999999999999999999

%N a(n) = 5*10^n - 1.

%C Also maximal value of GCD of 2 distinct (n+1)-digit numbers (compare with A126687). - _Michel Marcus_, Jun 24 2013

%C Also, a(n) is the largest obtained remainder when an (n+1)-digit number m is divided by any k with 1 <= k <= m. This remainder is obtained when 10^(n+1)-1 is divided by 5*10^n, example: 999 = 500 * 1 + 499, and a(2) = 499. - _Bernard Schott_, Nov 23 2021

%C Also numbers k whose digital reversal equals 2*(k - 2). - _Stefano Spezia_, Sep 15 2024

%H Vincenzo Librandi, <a href="/A198971/b198971.txt">Table of n, a(n) for n = 0..200</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-10).

%F a(n) = 10*a(n-1) + 9.

%F a(n) = 11*a(n-1) - 10*a(n-2), n>1.

%F G.f.: (4 + 5*x)/(1 - 11*x + 10*x^2). - _Vincenzo Librandi_, Jan 03 2013

%F E.g.f.: exp(x)*(5*exp(9*x) - 1). - _Stefano Spezia_, Nov 17 2022

%t CoefficientList[Series[(4 + 5*x)/(1 - 11*x + 10*x^2), {x, 0, 30}], x] (* _Vincenzo Librandi_, Jan 03 2013 *)

%t LinearRecurrence[{11,-10},{4,49},20] (* _Harvey P. Dale_, Dec 30 2018 *)

%o (Magma) [5*10^n-1 : n in [0..20]];

%o (PARI) a(n)=5*10^n-1 \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Cf. A002283, A126687.

%K nonn,easy

%O 0,1

%A _Vincenzo Librandi_, Nov 02 2011