OFFSET
0,3
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]
LINKS
David A. Corneth, Table of n, a(n) for n = 0..1000
FORMULA
a(p) = 3*p for all primes p >= 7.
EXAMPLE
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.
CROSSREFS
KEYWORD
nonn
AUTHOR
Peter Kagey, Nov 19 2019
EXTENSIONS
a(42)-a(43) and a(45) from David A. Corneth, Dec 25 2021
More terms from Jinyuan Wang, Dec 26 2021
STATUS
approved