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A061491
a(1) = 1, a(n) = least number such that the concatenation a(n)a(n-1)...a(1) is a cube.
2
1, 133, 100330363, 100000000300330000300660363, 100000000000000000000000000300000000300330000000000000300000000600660000300660363
OFFSET
1,2
LINKS
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
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