A061491 a(1) = 1, a(n) = least number such that the concatenation a(n)a(n-1)...a(1) is a cube.

1, 133, 100330363, 100000000300330000300660363, 100000000000000000000000000300000000300330000000000000300000000600660000300660363

Offset: 1

Author

Amarnath Murthy, May 06 2001

Keywords

• base
• nonn

Examples

			a(1) = 1, a(2) = 133, a(2)a(1) = 1331 = 11^3.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..7

Crossrefs

Cf. A061359, A061361, A061363.

Programs

Maple

f:= proc(n) local j; 10^(-3^(n-1)/2-1)*(add(10^(3^j/2),j=0..n-1)^3 - add(10^(3^j/2),j=0..n-2)^3) end proc:
seq(f(n),n=1..5); # Robert Israel, Jun 17 2024

Formula

a(n) = 10^(-3^(n-1)/2-1)*(Sum_{j=0..n-1}(10^(3^j/2))^3 - Sum_{j=0..n-2} (10^(3^j/2))^3). - Robert Israel, Jun 17 2024

Extensions

More terms from Ulrich Schimke (ulrschimke(AT)aol.com), Nov 06 2001

Corrected by Robert Israel, Jun 17 2024