A338137 -id:A338137 - cp's OEIS frontend

A338138 Nested cube root (b(n) + (b(n-1) + ... + (b(1))^(1/3)...)^(1/3))^(1/3), where b(n) = A338137(n).

1, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 3, 3, 4, 3, 5, 3, 6, 3, 7, 4, 4, 5, 4, 6, 4, 7, 5, 5, 6, 5, 7, 6, 6, 7, 7, 8, 3, 8, 4, 8, 5, 8, 6, 8, 7, 9, 3, 9, 4, 9, 5, 9, 6, 9, 7, 10, 3, 10, 4, 10, 5, 10, 6, 10, 7, 11, 3, 11, 4, 11

Offset: 1

Vincent Chan, Oct 12 2020

nonn

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.

Cf. A338137

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)

a(n) = (a(n-1) + A338137(n))^(1/3). - Andrew Howroyd, Oct 15 2020

cp's OEIS Frontend

A338138 Nested cube root (b(n) + (b(n-1) + ... + (b(1))^(1/3)...)^(1/3))^(1/3), where b(n) = A338137(n).

Views

Author

Keywords

Comments

Crossrefs

Programs

Python

Formula