A050647 a(n+1)^3 is next smallest cube ending with a(n)^3, initial term is 1.

1, 11, 10011, 10000000010011, 10000000000000000000000000010000000010011

Offset: 1

Author

Patrick De Geest, Jul 15 1999

Keywords

• nonn
• base

Comments

Frank Ellermann conjectures (Jun 11 2001) 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.

a(6) is too large to include (contains 122 digits).

Crossrefs

Cf. A050646, A048564, A050641.

Formula

a(n) = Sum_{i=0..n-1} 10^((3^i-1)/2). - Max Alekseyev, Jan 07 2015

Extensions

More terms from Ulrich Schimke (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