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%I #6 Apr 29 2019 20:37:17
%S 0,3,9,84,4041,5673648,48020423368761,806086788756824484462571572,
%T 221815145293562950532110825781341443907408910699844537
%N a(0) = 0, a(1) = 3; a(n) = rev(a(n-1))*a(n-1) + a(n-2), where rev = digit reversal (A004086).
%C The next term is too large to include.
%e +---+--------------+---------------------+------------------+
%e | n | a(n)/a(n+1) | Continued fraction | Comment |
%e +---+--------------+---------------------+------------------+
%e | 1 | 3/9 | [0; 3] | 3 = rev(a(1)) |
%e +---+--------------+---------------------+------------------+
%e | 2 | 9/84 | [0; 9, 3] | 9 = rev(a(2)) |
%e +---+--------------+---------------------+------------------+
%e | 3 | 84/4041 | [0; 48, 9, 3] | 48 = rev(a(3)) |
%e +---+--------------+---------------------+------------------+
%e | 4 | 4041/5673648 | [0; 1404, 48, 9, 3] | 1404 = rev(a(4)) |
%e +---+--------------+---------------------+------------------+
%t a[n_] := a[n] = FromDigits[Reverse[IntegerDigits[a[n - 1]]]] a[n - 1] + a[n - 2]; a[0] = 0; a[1] = 3; Table[a[n], {n, 0, 8}]
%Y Cf. A001040, A004086, A058182.
%K nonn,base
%O 0,2
%A _Ilya Gutkovskiy_, Apr 29 2019
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