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A338137
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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.
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1
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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PROG
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(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)
(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
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CROSSREFS
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Cf. A323635 (similar definition with square roots).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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