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Numbers k such that A163511(k) is a cube.
6

%I #16 Oct 02 2023 12:51:14

%S 0,4,9,19,32,39,65,72,79,131,145,152,159,256,263,291,305,312,319,513,

%T 520,527,576,583,611,625,632,639,1027,1041,1048,1055,1153,1160,1167,

%U 1216,1223,1251,1265,1272,1279,2048,2055,2083,2097,2104,2111,2307,2321,2328,2335,2433,2440,2447,2496,2503,2531,2545,2552

%N Numbers k such that A163511(k) is a cube.

%C The sequence is defined inductively as:

%C (a) it contains 0 and 4,

%C and

%C (b) for any nonzero term a(n), (2*a(n)) + 1 and 8*a(n) are also included as terms.

%C Because the inductive definition guarantees that all terms after 0 are of the form 7k+2, 7k+4 or 7k+5 (A047378), and because for any n >= 0, n^3 == 0, 1 or 6 (mod 7), (i.e., cubes are in A047275), it follows that there are no cubes in this sequence after the initial 0.

%o (PARI)

%o A163511(n) = if(!n, 1, my(p=2, t=1); while(n>1, if(!(n%2), (t*=p), p=nextprime(1+p)); n >>= 1); (t*p));

%o isA365801(n) = ispower(A163511(n),3);

%o (PARI) isA365801(n) = if(n<=4, !(n%4), if(n%2, isA365801((n-1)/2), if(n%8, 0, isA365801(n/8))));

%Y Positions of multiples of 3 in A365805.

%Y Sequence A243071(n^3), n >= 1, sorted into ascending order.

%Y Subsequence of A047378 (after the initial 0).

%Y Subsequences: A013731, A153894.

%Y Cf. A000578, A047275, A163511.

%Y Cf. also A365802, A365808.

%K nonn

%O 1,2

%A _Antti Karttunen_, Oct 01 2023