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a(1) = 1, a(n+1) is the numerator of n-th partial fraction of the continued fraction [1; 10, 100, 1000, ..., 10^n].
1

%I #33 Dec 31 2016 06:16:31

%S 1,11,1101,1101011,11010111101,1101011111201011,

%T 1101011111212021111101,11010111112121312122121201011,

%U 1101011111212132313223231313121111101,1101011111212132324233342425242423223121201011

%N a(1) = 1, a(n+1) is the numerator of n-th partial fraction of the continued fraction [1; 10, 100, 1000, ..., 10^n].

%H Seiichi Manyama, <a href="/A280042/b280042.txt">Table of n, a(n) for n = 1..45</a>

%F a(n) = 10^(n-1)*a(n-1) + a(n-2).

%e G.f. = x + 11*x^2 + 1101*x^3 + 1101011*x^4 + 11010111101*x^5 + 1101011111201011*x^6 + ...

%e a(3) = 1101, the numerator of 1 + 1/(10 + 1/100) = 1101/1001.

%Y Denominators are in A015482.

%Y Cf. A061377, A280219, A280220.

%K nonn,frac

%O 1,2

%A _Seiichi Manyama_, Dec 31 2016