A338137 Lexicographically earliest sequence of distinct positive integers such that the nested cube root (a(n) + (a(n-1) + ... + (a(1))^(1/3)...)^(1/3))^(1/3) is an integer.

1, 7, 6, 25, 5, 62, 4, 123, 3, 214, 2, 341, 20, 24, 61, 23, 122, 22, 213, 21, 340, 57, 60, 121, 59, 212, 58, 339, 118, 120, 211, 119, 338, 209, 210, 337, 336, 505, 19, 509, 56, 508, 117, 507, 208, 506, 335, 722, 18, 726, 55, 725, 116, 724, 207, 723, 334, 993, 17, 997, 54, 996

Offset: 1

Author

Vincent Chan, Oct 12 2020

Keywords

• nonn

Comments

A permutation of positive integers: letting s_n = (a(n) + (a(n-1) + ... + (a(1))^(1/3)...)^(1/3))^(1/3), we have a_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 a_1,...,a_{n-1} <= k^3. Since (k+1)^3-s_{n-1} >= k^3+3k^2+2k+1 > a_j for j < n and a_n is the smallest positive integer not already in the sequence for which a_n+s_{n-1} is a cube, then s_n <= k+1. Then we note that a_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. Finally, for an integer k, k appears in the sequence {a_n} no later than the first time s_{n-1} = k^3-k.

Crossrefs

Cf. A323635 (similar definition with square roots).

Programs

PARI

lista(nn) = {my(va = vector(nn), lastcb); va[1] = 1; lastcb = 1; for (n=2, nn, my(k = ceil(sqrtn(sqrtnint(lastcb, 3), 3))); while (#select(x->(x==(k^3-sqrtnint(lastcb, 3))), va), k++); va[n] = k^3-sqrtnint(lastcb, 3); lastcb = k^3;); va; } \\ Michel Marcus, Oct 13 2020

Python

myList = [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
            t = 1
        else:
            b += 1
    t=0
print("myList: ",myList)