%I #24 Apr 11 2022 17:02:24
%S 0,1,4,9,9,18,18,21,8,16,24,33,18,39,28,30,25,51,25,57,36,36,44,69,42,
%T 36,52,27,45,87,45,93,49,55,68,60,48,111,76,65,60,123,54,129,66,54,92,
%U 141,70,56,72,85,78,159,80,80,84,95,116,177,84,183,124,84,64
%N a(n) is the smallest m for which there is a sequence n = b_1 < b_2 < ... < b_t = m such that b_1*b_2*...*b_t is a perfect cube.
%C For each k there exists a sufficiently large N such that for all primes p > N, a(k*p) = (k+2)*p. [We can prove the proposition is true for N = 64*(t*k)^2, where t = k*(k+1)*(k+2): there is a positive integer x such that t^2*x^3 < k*p < t^2*(x+1)^3 < t^2*x^3*(k+1)/k < (k+1)*p for p > N. So one increasing sequence starting with k*p, ending with (k+2)*p, and having a product which is a perfect cube is (k*p) * (t^2*(x+1)^3) * ((k+1)*p) * ((k+2)*p) = (t*p*(x+1))^3. Noticed that a(k*p) >= (k+2)*p (because b_1*b_2*...*b_t is divisible by p^3) for p > N, so a(k*p) = (k+2)*p. - _Jinyuan Wang_, Dec 22 2021]
%H David A. Corneth, <a href="/A329732/b329732.txt">Table of n, a(n) for n = 0..1000</a>
%F a(p) = 3*p for all primes p >= 7.
%e For n = 22, one increasing sequence starting with 22, ending with a(22) = 44, and having a product which is a perfect cube is 22 * 24 * 25 * 30 * 32 * 33 * 44 = 2640^3.
%Y A cube analog of R. L. Graham's sequence (A006255).
%Y Cf. A277494.
%K nonn
%O 0,3
%A _Peter Kagey_, Nov 19 2019
%E a(42)-a(43) and a(45) from _David A. Corneth_, Dec 25 2021
%E More terms from _Jinyuan Wang_, Dec 26 2021