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Record values occur at prime values of n, and equal one less than the next lowest prime number (see Formula). Because of this, a(n) is always less than n, so for any positive integer starting value n, iterations of n -> n + gpf(n) will eventually join A076271.

If a(k) is the first occurrence of a prime p(r), the r-th prime then a(k+1) = p(r-1) + 1,... a(k+m) = p(r-1) + m until a(k + p(r)-p(r-1))= p(r) again.( the second and the last occurrence of p(r)).

Or, for n>1, n + (largest prime divisor of n). [Anne Robinson, daughter of Herman P. Robinson, Oct 08 1981]

After the initial term, each even-indexed term equals the preceding term plus its largest prime factor, and each odd-indexed term equals the preceding term plus its smallest prime factor.

See also sequence A076271 where a(n+1) = a(n) + lpf(a(n)).

Each term shares exactly one prime factor with the immediately preceding term, and because the sequence is strictly increasing, all the terms after 2 are composite. - Antti Karttunen, Apr 19 2015

From a(3) onward, the terms are alternately even and odd. - Jan Guichelaar, Apr 24 2015

a(2*n) = A070229(a(2*n-1)); a(2*n+1) = A061228(a(2*n)). - Reinhard Zumkeller, May 06 2015

For prime p let [p] denote the sequence with a(1)=p, and generated as for the terms of the current sequence (which according to this notation is then the same as [2]). It so happens that the sequence [p] (for any p?) merges with [2] sooner or later, taking the form of a "tree" as shown in the attached image (Including prime starts up to p=67). Is this pattern of merging bounded or not? Is there just one tree or are there many? Interesting to speculate. The numbers corresponding to the arrival points in [2] of [p] is the sequence 2,6,15,21,51,57,77,84.... The sequence of ("excluded") numbers which do not arise in [p] for any prime p starts as 8,16,20,25,28,32,36,44... Other sequences may refer to the number of iterations required to merge [p] into [2]. See tree picture. - David James Sycamore, Aug 25 2016

In this picture, one could also include some [c] sequences, with composite c, see A276269. - Michel Marcus, Aug 26 2016

a(n) satisfies the following inequality: (1/4)*(n^2 + 3*n + 1) <= a(n) <= (1/4)*(n+2)^2. [Corrected by M. F. Hasler, Apr 08 2015]

The present sequence is the special case a(n) = a(2,n) with a more general a(m, n) := a(m, n-1) + gpf(a(m, n-1)), a(m, 1) := m, where gpf(x) := "greatest prime factor of x" = A006530(x). Also a(a(r,k), n) = a(r,n+k-1), for all n,k in N\{0} and all r in N\{0,1}; a(prime(k), n) = a(prime(i), n + prime(k) - prime(i)), for all k,i,n in N\{0}, with k >= i, n >= prime(k-1) and with prime(x) := x-th prime.

Essentially the same as A076271 and A180107, cf. formula.

Apart from initial terms, same as A036441 and A076271, but the simple definition warrants an independent entry.

Conjecture: with the requirement that the prime factorization of each term is written so that the primes are ordered from smallest to largest, the sequence is the lexicographically earliest infinite sequence of distinct positive numbers such that gpf(a(n-1)) * lpf(a(n)) = |a(n) - a(n-1)|, where gpf(k) = A006530(k) = greatest prime factor of k and lpf(k) = A020639(k) = least prime factor of k. In this way the sequence is the ordered prime factorization version of the 'Commas sequence', A121805. One can show that, for such a sequence to be infinite, no odd number can appear. Although for many terms a lower even number can be chosen for the following term, which can lead to even lower numbers for further terms, it is conjectured all such choices will ultimately halt the sequence as a number is eventually reached for which no unused next number exists which follows the required rule for the difference between the terms. Therefore all terms must be larger than the previous, and the earliest such infinite sequence is the given sequence.

See A367465 for the sequence when the requirement that the primes in the factorization of each term must be in order is removed.