cp's OEIS Frontend

For all n, a(n) <> A335882(n). Proof: We need to consider only the odd terms, because for n > 1, A332215(2^k * n) = 2^k * A332215(n). The odd terms of A335882 are either primes or semiprimes whose both factors are Mersenne primes, terms of A144482.

(A) If A335882(n) is a prime, then a(n) = A332215(A335882(n)) is a term of A000225 (of the form 2^k - 1, a binary repunit), while primes in A335882 are certainly not of that form, as all Mersenne primes (A000668) are on a different row in array A335430 (on row 1, A335431).

(B) For any semiprime k in A335882, there is only one non-leading zero in the binary representation of A332215(k). On the other hand, a product of two Mersenne primes always contains more than one non-leading zero in its base-2 representation: for three times a Mersenne prime, there are two such zeros, as explained in A279389, and products of two Mersenne primes > 3 are always of the form 8k+1, with at least two zeros immediately left of the least significant 1-bit.

Let f(n) = A000265(n) be the odd part of n. Let p be the largest prime factor of k, and say k = p * m. Suppose that k is not a power of 2, i.e., p > 2, then f(k) = p * f(m). The iteration is k -> k + k/p = p*m + m = (p+1) * m. So, p * f(m) -> f(p+1) * f(m). Since for p > 2, f(p+1) < p, the odd part in each iteration decreases, until it becomes 1, i.e., until we reach a power of 2. - Amiram Eldar, Feb 19 2020

Any odd prime factor of k can be used at any step of the iteration, and the result will be same. Thus, like A329697, this is also fully additive sequence. - Antti Karttunen, Apr 29 2020

If and only if a(n) is equal to A005087(n), then sigma(2n) - sigma(n) is a power of 2. (See A336923, A046528). - Antti Karttunen, Mar 16 2021

Numbers of the form 2^k * ((2^p)-1), where p is one of the primes in A000043, and k >= 0.

Numbers k such that A000265(k) is in A000668.

Numbers k for which A331410(k) = 1.

Numbers k that themselves are not powers of two, but for which A335876(k) = k+A052126(k) is [a power of 2].

Conjecture: This sequence gives all fixed points of map n -> A332214(n) and its inverse n -> A332215(n). See also notes in A029747 and in A163511.

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)

As the product of any two primes is semiprime by definition, this is also the list of composite numbers n=x*y where both x and y are Mersenne primes. - Christian N. K. Anderson, Mar 25 2013

Array is read by descending antidiagonals with (n,k) = (0,0), (0,1), (1,0), (0,2), (1,1), (2,0), ... where A(n,k) is the (k+1)-th solution x to A331410(x) = n. The row indexing (n) starts from 0, and column indexing (k) also from 0.

For any odd prime p that appears on row n, p+1 appears on row n-1.

The e-th powers of the terms on row n form a subset of terms on row (e*n). More generally, a product of terms that occur on rows i_1, i_2, ..., i_k can be found at row (i_1 + i_2 + ... + i_k), because A331410 is completely additive.

Primes p such that A331410(1+p) = 1. After 2, primes p for which A331410(p) = 2.

Numbers n such that when we start from k = n, and apply in some combination the nondeterministic maps k -> k - k/p and k -> k + k/p, (where p can be any of the odd prime factors of k), then for some combination we can reach a power of 2 in exactly two steps (but with no combination allowing 0 or 1 steps).

Numbers n such that when we start from k = n, and apply in any combination the nondeterministic maps k -> k - k/p and k -> k + k/p, (where p can be any of the odd prime factors of k), a power of 2 will appear no later than after two such steps, and on some of the combinations a power of 2 will appear after exactly two steps.