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If in the search tree (whose root is n) where the transitions x -> A003415(x) and x -> A276086(x) expand the tree at every point to two branches, the latter transition yields a number larger than A002110(37), that number (and the whole subtree whose root it is) will then be pruned from the tree, which entails that the procedure could miss shorter paths to zero than what it will find later from the other branches of the tree. Thus for all n, this sequence gives the guaranteed upper bound for the terms of A327969.

Note that A002110(37) = 35375166993717494840635767087951744212057570647889977422429870 is a 205-bit integer.

Prime product form of primorial base expansion of n.

Sequence is a permutation of A048103. It maps the smallest prime not dividing n to the smallest prime dividing n, that is, A020639(a(n)) = A053669(n) holds for all n >= 1.

The sequence satisfies the exponential function identity, a(x + y) = a(x) * a(y), whenever A329041(x,y) = 1, that is, when adding x and y together will not generate any carries in the primorial base. Examples of such pairs of x and y are A328841(n) & A328842(n), and also A328770(n) (when added with itself). - Antti Karttunen, Oct 31 2019

From Antti Karttunen, Feb 18 2022: (Start)

The conjecture given in A327969 asks whether applying this function together with the arithmetic derivative (A003415) in some combination or another can eventually transform every positive integer into zero.

Another related open question asks whether there are any other numbers than n=6 such that when starting from that n and by iterating with A003415, one eventually reaches a(n). See comments in A351088.

This sequence is used in A351255 to list the terms of A099308 in a different order, by the increasing exponents of the successive primes in their prime factorization. (End)

From Bill McEachen, Oct 15 2022: (Start)

From inspection, the least significant decimal digits of a(n) terms form continuous chains of 30 as follows. For n == i (mod 30), i=0..5, there are 6 ordered elements of these 8 {1,2,3,6,9,8,7,4}. Then for n == i (mod 30), i=6..29, there are 12 repeated pairs = {5,0}.

Moreover, when the individual elements of any of the possible groups of 6 are transformed via (7*digit) (mod 10), the result matches one of the other 7 groupings (not all 7 may be seen). As example, {1,2,3,6,9,8} transforms to {7,4,1,2,3,6}. (End)

The least significant digit of a(n) in base 4 is given by A353486, and in base 6 by A358840. - Antti Karttunen, Oct 25 2022, Feb 17 2024

The first derivative of 0 and 1 is 0. The second derivative of a prime number is 0.

For all n, A003415(a(n)) is also a term of the sequence. A351255 gives the nonzero terms as ordered by their position in A276086. - Antti Karttunen, Feb 14 2022

Equivalently, solutions to n'' = 1, since n' = 1 iff n is prime. Twice the lesser of the twin primes, 2*A001359 = A108605, are a subsequence. - M. F. Hasler, Apr 07 2015

All terms are squarefree, because if there would be a prime p whose square p^2 would divide n, then A003415(n) = (A003415(p^2) * (n/p^2)) + (p^2 * A003415(n/p^2)) = p*[(2 * (n/p^2)) + (p * A003415(n/p^2))], which certainly is not a prime. - Antti Karttunen, Oct 10 2019

Under iterations of the arithmetic derivative, the orbit of some numbers ends in zero, and the orbit of all others (I conjecture) reaches a number of the form m*p^p with prime p, from where on it keeps this form and grows to infinity iff m>1, or remains at this fixed point if m=1.

This is an extension of the sequence A099307 which counts the steps to reach 0 or yields 0 if this never happens.

Numbers n such that A327859(n) = A276086(A003415(n)) is an odd prime.

Composite terms in A328232.

Although it first might seem that the numbers whose arithmetic derivative is A002110(k) all appear before any of those whose arithmetic derivative is A002110(k+1), that is not true, as for example, we have a(56) = 570149, and A003415(570149) = 2310, a(57) = 570209, and A003415(570209) = 30030, but then a(58) = 573641 with A003415(573641) = 2310 again.

Because this is a subsequence of A327862 (all primorials > 1 are of the form 4k+2), only odd numbers are present.

Conjecture: No multiples of 5 occur in this sequence, and no multiples of 3 after the initial 9.

Of the first 10000 terms, all others are semiprimes (with 9 the only square one), except 1547371 = 7^2 * 23 * 1373 and 79332523 = 17^2 * 277 * 991, the latter being the only known term whose decimal expansion ends with 3. If all solutions were semiprimes p*q such that p+q = A002110(k) for some k > 1 (see A002375), it would be a sufficient reason for the above conjecture to hold. - David A. Corneth and Antti Karttunen, Oct 11 2019

In any case, the solutions have to be of the form "odd numbers with an even number of prime factors with multiplicity" (see A235992), and terms must also be cubefree (A004709), as otherwise the arithmetic derivative would not be squarefree.

Sequence A366890 gives the non-Goldbachian solutions, i.e., numbers that are not semiprimes. See also A368702. - Antti Karttunen, Jan 17 2024

Numbers x such that A276086(x) [which is A351255(a(x))] is in A099308.

The fourth A003415-iteration of a(n) is the first to be 0.