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

1,2

Links

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

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

Example

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

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

Crossrefs

Cf. A061359, A061361, A061363.

Sequence in context: A055579 A191715 A208626 * A274132 A252133 A255795

Adjacent sequences: A061488 A061489 A061490 * A061492 A061493 A061494

Keyword

base,nonn

Author

Amarnath Murthy, May 06 2001

Extensions

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

Corrected by Robert Israel, Jun 17 2024

Status

approved