cp's OEIS Frontend

Without taking the floor there is no such Q where (R-Q)*(Q-P)/8 = 2.

Dickson's conjecture implies that if n == 0 or 1 (mod 3) there are infinitely many primes Q such that Q - 2 is the previous prime and Q + 4 n is the next prime, and that if n == 2 (mod 3) there are infinitely many primes Q such that Q - 2 is the previous prime and Q + 4 n + 2 is the next prime. - Robert Israel, Sep 24 2024

Start with all terms set to the nominal value a(n) = n.

Mark all a(n), n > 1 as "unedited", and set a pointer p = 1.

For each unedited a(n), in increasing order of n, let q = n + a(p). Look at a(q). If a(q) has previously been edited, set a(n) = a(q); otherwise, set a(n) = q. In either case, then set a(q) = n. Mark both a(n) and a(q) as "edited". Increase p by 1 and move to the next unedited a(n), etc.

Note that once the process reaches any particular unedited n, sequence terms a(1) to a(n-1) will not be edited further by the generation process and are thus fixed. Terms of the sequence for entries greater than n may have been edited, but are "in flux" and may change again until n reaches them. p necessarily lags behind n, and so any a(p) is always well fixed by the time it is needed.

In many cases the generation process merely swaps two values relative to the positive integers, but since some edited terms could be re-edited, this will cause cycles to occur.

The positions of integers in this sequence would appear to be very hard to predict without actually generating the sequence.

Permutations in which adjacent terms sum to a particular value is a property central to the sequences A090460, A071984, A108658, A272259, and A107929.

0 and 1 are not prime and are single digits in all bases, so no reversal of digits can make them prime. a(n) is therefore 0 for both.

4 is not prime and so cannot be prime if reversed in any base where it is a single digit. This leaves bases 2 and 3 where, upon reversal, it is 1 and 4 respectively. Neither are prime, and so a(4) is also 0.

Conjecture 1: 0, 1 and 4 are the only values where there is no base in which a digital reversal makes a prime.

It is clear that for any prime p, a(p) cannot be zero, since a(p)=p+1 is a solution if there is none smaller.

Conjecture 2: n = 2 is the only prime p which must be represented in base p+1, i.e., trivially, as a single digit, in order for its reversal to be prime.

Corollary: Since a(n) cannot be n itself -- reversing n in base n obtains 1, which is not prime -- this would mean that for all positive n except 2, a(n) < n.

Other than its small magnitude, a(n) = 2 occurs often due to the fact that a reversed positive binary number is guaranteed to be odd and thus stands a greater chance of being prime.

Similarly, many solutions exist solely because reversal removes all powers of the base from n, reducing the number of divisors. Thus based solely on observation:

Conjecture 3: With the restriction gcd(base,n) = 1, a(n) = 0 except for n = 2, 3 and 6k+-1, for positive integer k, i.e., terms of A038179.

An enumeration of all sorted k-tuples containing positive integers.

Begin with the 1-tuple (1), and then reading from the beginning of the list of k-tuples append to the list (n+1) if the k-tuple read is a 1-tuple and for all cases, append the (k+1)-tuples (1,n,...), (2,n,...), ..., (n,n,...), where n is the first element of the k-tuple that was read.

This sequence is a flattening of that process.

Other properties of this sequence are as A229874.

The two ones at the start of the parent sequence represent parent and child 1-tuples in the grandparent sequence [(1) and (2) respectively], hence this sequence also starts with 1, 1 rather than 2, which would otherwise be a more sensible way to describe the pair of ones.

All other elements are effectively run-lengths of strings of the same integer in A229895.

The first occurrence of an integer, n, in the parent sequence, is the first of a run of n^n elements of value n. For later occurrences, the run length is n^k-(n-1)^k where k is the size of the k-tuple in the grandparent sequence, A229873.

The elements can be arranged into a triangle thus:

.... 1

.... 1, 4

.... 1, 5, 27

.... 1, 7, 37, 256

.... 1, 9, 61, 369, 3125

.... etc.

where the n-th line is:

.... n^1-(n-1)^1, n^2-(n-1)^2, ..., n^(k-1)-(n-1)^(k-1), n^n; 1 <= k < n

The first terms, for sufficiently large n simplifying as:

.... 1, 2n-1, 3n^2-3n+1, 4n^3-6n^2+4n-1, etc.

Row sums are first differences of A031972, and thus the cumulative sum of rows at the end of each row is A031972 itself, i.e., n*(n^n - 1)/(n-1).

Number k of elements to read from A229873 to obtain the next k-tuple.

The sequence pattern is an integer, n, followed by all k-tuples containing n, then (k+1)-tuples, etc., up to the n-tuples that have not yet appeared in the sequence. Directly before the integer n+1, therefore, we find the first occurrence of n^n n-tuples which contain the n^n permutations of 1 to n in lexicographic order. The cases n = 1 and n = 2 are degenerate as no tuples precede them; 1 is followed not by a tuple, but by 2, and 2 is followed by the tuple (1, 1), rather than (1, n) as with all other integers.

k-tuple clusters later in the sequence (k

Essentially, at each stage an n-hypercube of elements of size n is completed for each dimension up to the (n-1)-th, building on previous occurrences of the dimension, and then a hypercube for dimension n is begun to be built upon later.

Tuple sizes are in A229895.