cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

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%I A338137 #12 Oct 15 2020 17:01:25
%S A338137 1,7,6,25,5,62,4,123,3,214,2,341,20,24,61,23,122,22,213,21,340,57,60,
%T A338137 121,59,212,58,339,118,120,211,119,338,209,210,337,336,505,19,509,56,
%U A338137 508,117,507,208,506,335,722,18,726,55,725,116,724,207,723,334,993,17,997,54,996
%N 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.
%C A338137 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.
%o A338137 (Python)
%o A338137 myList = [1]
%o A338137 s = 1
%o A338137 t = 0
%o A338137 for n in range(9999):
%o A338137     b = 2
%o A338137     while t == 0:
%o A338137         if(b**3-s > 0 and not b**3-s in myList):
%o A338137             myList.append(b**3-s)
%o A338137             s = b
%o A338137             t = 1
%o A338137         else:
%o A338137             b += 1
%o A338137     t=0
%o A338137 print("myList: ",myList)
%o A338137 (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
%Y A338137 Cf. A323635 (similar definition with square roots).
%K A338137 nonn
%O A338137 1,2
%A A338137 _Vincent Chan_, Oct 12 2020