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EXTENSIONS
| Jun 11, 2001: Frank Ellermann conjectures that if b(n) = 10^(3^n -1), i.e. 1, 100, 100000000, etc. then a(n) = concatenation b( n-2 ) || a( n-1 ) for n > 1.
More terms from UlrSchimke(AT)aol.com, Nov 06, 2001, who remarks that a(6) = 10^121+a(5) = concat (10^80, a(5)) and a(7) = 10^364+a(6) = concat (10^242, a(6)), which supports Ellermann's conjecture.
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