OFFSET
1,2
COMMENTS
Contains every positive integer: letting s_n = (b(n) + (b(n-1) + ... + (b(1))^(1/3)...)^(1/3))^(1/3), we have b_n = s_n^3-s_{n-1}. We claim that if s_1,...s_{n-1} <= k, then s_n <=k+1. Indeed, the given condition implies b_1,...,b_{n-1} <= k^3. Since (k+1)^3-s_{n-1} >= k^3+3k^2+2k+1 > b_j for j < n and b_n is the smallest positive integer not already in the sequence for which b_n+s_{n-1} is a cube, then s_n <= k+1. Then we note that b_n = s_n^3-s_{n-1} cannot repeat, so that s_n cannot be a single constant infinitely often, so {s_n} contains every positive integer.
FORMULA
a(n) = (a(n-1) + A338137(n))^(1/3). - Andrew Howroyd, Oct 15 2020
PROG
(Python)
myList = [1]
roots = [1]
s = 1
t = 0
for n in range(9999):
b = 2
while t == 0:
if(b**3-s > 0 and not b**3-s in myList):
myList.append(b**3-s)
s = b
roots.append(s)
t = 1
else:
b += 1
t=0
print("roots: ", roots)
CROSSREFS
KEYWORD
nonn
AUTHOR
Vincent Chan, Oct 12 2020
STATUS
approved