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%I #37 Apr 23 2024 16:28:49
%S 1,19,109,1099,10099,100999,1000999,10009999,100009999,1000099999,
%T 10000099999,100000999999,1000000999999,10000009999999,
%U 100000009999999,1000000099999999,10000000099999999,100000000999999999,1000000000999999999,10000000009999999999,100000000009999999999
%N a(n) is the n-digit number k such that R(k)/k is maximal for any n-digit number.
%H Michael S. Branicky, <a href="/A372010/a372010.txt">Proof of form of a(n) for OEIS A372010</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (11,0,-110,100).
%F a(n) = 1 0^(n-1-h) 9^h, where h = floor(n/2) and ^ represents repeated concatenation (see links for proof).
%F a(n) = 10^(n-1) + 10 ^ floor(n / 2) - 1.
%F G.f.: x*(1 + 8*x - 100*x^2 + 10*x^3)/((1 - x)*(1 - 10*x)*(1 - 10*x^2)). - _Stefano Spezia_, Apr 16 2024
%e a(2) = 19 as k = 19 is the two digit number k that produces the largest ratio R(k)/k = 91/19 of all two-digit numbers.
%t Table[10^(n-1) + 10^Floor[n/2] - 1, {n, 25}] (* _Paolo Xausa_, Apr 23 2024 *)
%o (PARI) a(n) = 10^(n-1) + 10 ^ (n \ 2) - 1
%o (Python) def a(n): return 10**(n-1) + 10**(n//2) - 1
%Y Cf. A004086.
%K nonn,easy
%O 1,2
%A _David A. Corneth_ and _Michael S. Branicky_, Apr 15 2024
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