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a(n) is the cubic root of A175428(n) + A175428(n+1).
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%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