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A280340 a(n) = a(n-1) + 10^n * a(n-2) with a(0) = 1 and a(1) = 1.

%I A280340 #17 Feb 16 2025 08:33:39
%S A280340 1,1,101,1101,1011101,111111101,1011212111101,1112122222111101,
%T A280340 101122323232322111101,1112223344434333322111101,
%U A280340 1011224344546565545343322111101,111223345667777878776655443322111101,1011224455769911213121200887756443322111101
%N A280340 a(n) = a(n-1) + 10^n * a(n-2) with a(0) = 1 and a(1) = 1.
%C A280340 The Rogers-Ramanujan continued fraction is defined by R(q) = q^(1/5)/(1+q/(1+q^2/(1+q^3/(1+ ... )))). The limit of a(n)/A015468(n+2) is 10^(-1/5) * R(10).
%C A280340 a(n) has A004652(n+1) digits.  The last n digits are the same as the last n digits of a(n-1). - _Robert Israel_, Jan 12 2017
%H A280340 Seiichi Manyama, <a href="/A280340/b280340.txt">Table of n, a(n) for n = 0..62</a>
%H A280340 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Rogers-RamanujanContinuedFraction.html">Rogers-Ramanujan Continued Fraction</a>
%F A280340 a(n) a(n-3) = 10 a(n-2) a(n-1) - 10 a(n-2)^2 + a(n-1) a(n-3). - _Robert Israel_, Jan 12 2017
%e A280340 1/1 = a(0)/A015468(2).
%e A280340 1/(1+10/1) = 1/11 = a(1)/A015468(3).
%e A280340 1/(1+10/(1+10^2/1)) = 101/111 = a(2)/A015468(4).
%e A280340 1/(1+10/(1+10^2/(1+10^3/1))) = 1101/11111 = a(3)/A015468(5).
%p A280340 A[0]:= 1: A[1]:= 1:
%p A280340 for n from 2 to 20 do A[n]:= A[n-1]+10^n*A[n-2] od:
%p A280340 seq(A[i],i=0..20); # _Robert Israel_, Jan 12 2017
%t A280340 RecurrenceTable[{a[0]==a[1]==1,a[n]==a[n-1]+10^n a[n-2]},a,{n,15}] (* _Harvey P. Dale_, Jul 12 2020 *)
%Y A280340 Cf. A004652, A015468, A128915.
%Y A280340 Cf. similar sequences with the recurrence a(n-1) + q^n * a(n-2) for n>1, a(0)=1 and a(1)=1: A280294 (q=2), A279543 (q=3), this sequence (q=10).
%K A280340 nonn
%O A280340 0,3
%A A280340 _Seiichi Manyama_, Dec 31 2016