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A279543 a(n) = a(n-1) + 3^n * a(n-2) with a(0) = 1 and a(1) = 1.

%I A279543 #30 Feb 16 2025 08:33:38
%S A279543 1,1,10,37,847,9838,627301,22143007,4137864868,439978671649,
%T A279543 244776761262181,78185678507867584,130162592460442600405,
%U A279543 124783388108159412726037,622688428086038843429228482,1791127919536971393223950620041
%N A279543 a(n) = a(n-1) + 3^n * a(n-2) with a(0) = 1 and a(1) = 1.
%C A279543 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)/A015460(n+2) is 3^(-1/5) * R(3).
%H A279543 Seiichi Manyama, <a href="/A279543/b279543.txt">Table of n, a(n) for n = 0..90</a>
%H A279543 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Rogers-RamanujanContinuedFraction.html">Rogers-Ramanujan Continued Fraction</a>
%e A279543 1/1 = a(0)/A015460(2).
%e A279543 1/(1+3/1) = 1/4 = a(1)/A015460(3).
%e A279543 1/(1+3/(1+3^2/1)) = 10/13 = a(2)/A015460(4).
%e A279543 1/(1+3/(1+3^2/(1+3^3/1))) = 37/121 = a(3)/A015460(5).
%t A279543 RecurrenceTable[{a[n] == a[n - 1] + 3^n*a[n - 2], a[0] == 1, a[1] == 1}, a, {n, 15}] (* _Michael De Vlieger_, Dec 31 2016 *)
%Y A279543 Cf. A015460, A128915.
%Y A279543 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), this sequence (q=3), A280340 (q=10).
%K A279543 nonn
%O A279543 0,3
%A A279543 _Seiichi Manyama_, Dec 31 2016