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Numbers k such that A163511(k) is a fifth power.
6

%I #41 Jan 04 2024 10:55:50

%S 0,16,33,67,135,271,512,543,1025,1056,1087,2051,2113,2144,2175,4103,

%T 4227,4289,4320,4351,8207,8455,8579,8641,8672,8703,16384,16415,16911,

%U 17159,17283,17345,17376,17407,32769,32800,32831,33792,33823,34319,34567,34691,34753,34784,34815,65539,65601,65632,65663,67585,67616

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

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

%C The sequence is defined inductively as:

%C (a) it contains 0 and 16, and

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

%C 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].

%C 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.

%C In any case, this sequence do not contain any fifth powers after the initial zero. See A365805. - _Antti Karttunen_, Nov 23 2023

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

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

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

%Y Positions of multiples of 5 in A365805.

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

%Y Subsequences: A013825, A198275.

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

%Y Cf. also A365801, A365808, A366287, A366391.

%K nonn

%O 1,2

%A _Antti Karttunen_, Oct 01 2023