%I #11 Apr 29 2019 17:23:25
%S 2,3,4,5,6,7,6,3,4,5,6,7,6,3,4,5,6,7,6,3,4,5,6,7,6,3,4,5,6,7,6,3,4,5,
%T 6,7,6,3,4,5,6,7,8,7,4,5,6,5,2,3,4,5,6,7,6,2,3,4,5,6,7,8,9,8,5,4,5,6,
%U 7,8,7,5,6,7,6,4,5,6,7,8,9,10,9,6,5,6,7
%N a(n) is the cubic root of A175428(n) + A175428(n+1).
%C The sum of two consecutive terms of A175428 is always a perfect cube.
%H Rémy Sigrist, <a href="/A307785/b307785.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n)^3 = A175428(n) + A175428(n+1).
%e for n = 6:
%e - A175428(6) + A175428(7) = 135 + 208 = 343 = 7^3,
%e - hence a(6) = 7.
%o (PARI) p=1; s=0; for (n=1, 87, s+=2^p; for (v=1, oo, if (!bittest(s,v) && ispower(p+v, 3, &q), print1 (q ", "); p=v; break)))
%Y Cf. A175428.
%K nonn
%O 1,1
%A _Rémy Sigrist_, Apr 28 2019