|
%I #9 Jun 17 2024 15:25:33
%S 1,133,100330363,100000000300330000300660363,
%T 100000000000000000000000000300000000300330000000000000300000000600660000300660363
%N a(1) = 1, a(n) = least number such that the concatenation a(n)a(n-1)...a(1) is a cube.
%H Robert Israel, <a href="/A061491/b061491.txt">Table of n, a(n) for n = 1..7</a>
%F 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
%e a(1) = 1, a(2) = 133, a(2)a(1) = 1331 = 11^3.
%p 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:
%p seq(f(n),n=1..5); # _Robert Israel_, Jun 17 2024
%Y Cf. A061359, A061361, A061363.
%K base,nonn
%O 1,2
%A _Amarnath Murthy_, May 06 2001
%E More terms from Ulrich Schimke (ulrschimke(AT)aol.com), Nov 06 2001
%E Corrected by _Robert Israel_, Jun 17 2024
| 