A365802 Numbers k such that A163511(k) is a fifth power.

0, 16, 33, 67, 135, 271, 512, 543, 1025, 1056, 1087, 2051, 2113, 2144, 2175, 4103, 4227, 4289, 4320, 4351, 8207, 8455, 8579, 8641, 8672, 8703, 16384, 16415, 16911, 17159, 17283, 17345, 17376, 17407, 32769, 32800, 32831, 33792, 33823, 34319, 34567, 34691, 34753, 34784, 34815, 65539, 65601, 65632, 65663, 67585, 67616

Offset: 1

Author

Antti Karttunen, Oct 01 2023

Keywords

• nonn

Comments

Equivalently, numbers k for which A332214(k), and also A332817(k) are fifth powers.

The sequence is defined inductively as:

(a) it contains 0 and 16, and

(b) for any nonzero term a(n), (2*a(n)) + 1 and 32*a(n) are also included as terms.

When iterating n -> 2n+1 mod 31, starting from 16 we obtain five distinct remainders 16, 2, 5, 11, 23, before the cycle starts again from 16. (see A153893), while x^5 mod 31 may obtain only these values: 0, 1, 5, 6, 25, 26, 30. The only common element of these sets is 5. We have x^5 == 5 (mod 31) whenever x == 7, 14, 19, 25, 28 mod 31, with all other x leaving a remainder that is not in the set [16, 2, 5, 11, 23].

On the other hand, when iterating n -> 2n+1 mod 33, starting from 16 we obtain ten distinct remainders 16, 0, 1, 3, 7, 15, 31, 30, 28, 24, before the cycle starts again from 16, while x^5 mod 33 obtain only these values: 0, 1, 10, 11, 12, 21, 22, 23, 32. We have x^5 == 0 (mod 33) iff x == 0 (mod 33) and x^5 == 1 (mod 33) whenever x == 1, 4, 16, 25, 31 mod 33. In the n->2n+1 cycles of 5 and 10 elements starting from 16, the 5's (of every second cycle) in the former and the 1's in the latter are aligned with each other.

In any case, this sequence do not contain any fifth powers after the initial zero. See A365805. - Antti Karttunen, Nov 23 2023

Links

• Index entries for sequences related to binary expansion of n

Crossrefs

Positions of multiples of 5 in A365805.

Sequence A243071(n^5), n >= 1, sorted into ascending order.

Subsequences: A013825, A198275.

Cf. A000584, A153893, A163511, A332214, A332817.

Cf. also A365801, A365808, A366287, A366391.

Programs

PARI

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));
isA365802(n) = ispower(A163511(n),5);

PARI

isA365802(n) = if(n<=16, !(n%16), if(n%2, isA365802((n-1)/2), if(n%32, 0, isA365802(n/32))));