A329732 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.

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, 36, 52, 27, 45, 87, 45, 93, 49, 55, 68, 60, 48, 111, 76, 65, 60, 123, 54, 129, 66, 54, 92, 141, 70, 56, 72, 85, 78, 159, 80, 80, 84, 95, 116, 177, 84, 183, 124, 84, 64

Offset: 0

Author

Peter Kagey, Nov 19 2019

Keywords

• nonn

Comments

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]

Examples

			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.
		

Links

• David A. Corneth, Table of n, a(n) for n = 0..1000

Crossrefs

A cube analog of R. L. Graham's sequence (A006255).

Cf. A277494.

Formula

a(p) = 3*p for all primes p >= 7.

Extensions

a(42)-a(43) and a(45) from David A. Corneth, Dec 25 2021

More terms from Jinyuan Wang, Dec 26 2021