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Any Mersenne prime (A000668) times any power of 2 (i.e., 2^k, for k>=0) is fixed by this sequence, including also all even perfect numbers.

From Antti Karttunen, Jul 10 2020: (Start)

This is a "tuned variant" of A243071, and has many of the same properties.

For example, for n > 1, A007814(a(n)) = A007814(n) - A209229(n), that is, this map preserves the 2-adic valuation of n, except when n is a power of two, in which cases that value is decremented by one, and in particular, a(2^k * n) = 2^k * a(n) for all n > 1. Also, like A243071, this bijection maps primes to the terms of A000225 (binary repunits). However, the "tuning" (A332213) has a specific effect that each Mersenne prime (A000668) is mapped to itself. Therefore the terms of A335431 are fixed by this map. Furthermore, I conjecture that there are no other fixed points. For the starters, see the proof in A335879, which shows that at least none of the terms of A335882 are fixed.

(End)

Sequence is well-defined also in case there are only a finite number of Mersenne primes.

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

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

The sequence is defined inductively as:

(a) it contains 0 and 64,

and

(b) for any nonzero term a(n), (2*a(n)) + 1 and 128*a(n) are also included as terms.

When iterating n -> 2n+1 mod 127, starting from 64 we get 64, 2, 5, 11, 23, 47, 95, and then cycle starts again from 64 (see A153893), while on the other hand, x^7 mod 127 obtains values: 0, 1, 19, 20, 22, 24, 28, 37, 52, 59, 68, 75, 90, 99, 103, 105, 107, 108, 126. These sets have no terms in common, therefore there are no seventh powers in this sequence after the initial 0.

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

The sequence is defined inductively as:

(a) it contains 0 and 1024,

and

(b) for any nonzero term a(n), (2*a(n)) + 1 and 2048*a(n) are also included as terms.

When iterating n -> 2n+1 mod 2047, starting from 1024 we get 1024, 2, 5, 11, 23, 47, 95, 191, 383, 767, 1535, and then cycle starts again from 1024 (see A153893), while on the other hand, x^11 mod 2047 obtains values: 0, 1, 230, 322, 344, 368, 390, 482, 622, 712, 713, 942, 967, 1013, 1034, 1080, 1105, 1334, 1335, 1425, 1565, 1657, 1679, 1703, 1725, 1817, 2046. These sets have no terms in common, therefore there are no eleventh powers in this sequence after the initial 0.