cp's OEIS Frontend

A070229(n) is a lower bound of A006255(n) for all n in A102750.

This list contains all primes greater than 3 and no perfect squares.

Let k be a fixed integer, then k*p is found in this list for all sufficiently large primes p.

The powers of primes correspond to A000961.

Apparently, a(n) >= 0 for any n > 0.

For a given number n, while iterating A070229, we may encounter several prime increments (=several greatest prime factors). It is likely that the number of distinct increments before reaching a power of a prime is not bounded.

For k = 0..9, the least numbers with k distinct increments are:

k n Increments

- ------ ------------------------------------------

0 1 {}

1 6 {3}

2 12 {3, 5}

3 72 {3, 5, 17}

4 135 {5, 7, 11, 17}

5 686 {7, 11, 13, 19, 41}

6 12408 {47, 53, 59, 71, 89, 149}

7 35378 {19, 23, 67, 89, 101, 179, 211}

8 127581 {43, 53, 73, 103, 113, 227, 283, 457}

9 222111 {37, 79, 97, 191, 233, 239, 311, 359, 631}

The powers of primes correspond to A000961.

Apparently, a(n) > 0 for all n > 0.

60420 >= a(n) > 0 for n <= 6042, but either a(6043) = 0 or a(6043) > 10^30.

If p is prime, a(p) = a(2*p).

From n = 5, 55, 2, 24, 245, ... begin successive strings of exactly 1, 2, 3, 4, 5, ... identical merging points that are: 15, 105, 12, 77, 713, ... . - Bernard Schott, Jul 04 2020

Every nonprime appears exactly once in this sequence.

If n is a square we can take t=1 and a(n) = n. If n is a prime > 3, then a(n) = 2n and t=3. If n is twice a prime, say p, then a(n) = 3p most of the time. The sequence b_1 < b_2 < ... < b_t will not contain either perfect squares or primes for they bring nothing to the solution. Also I know of no n such that t = 2. - Robert G. Wilson v, Jan 30 2002

Let k be a fixed integer and p be a prime, then a(k*p) = (k+1)*p for sufficiently large p. - Peter Kagey, Feb 03 2015

From David A. Corneth, Oct 26 2016: (Start)

Is for all k*p in A277624, a(k*p) = (k+1) * p?

Conjecture: Let b(n) = A006530(A007913(n)). If b(n)^2 >= 2 * n then a(n) = n + b(n) except for n = 3, 10, and 171.

(End)

a(n) <= A072905(n).

a(n) <= 2*n for all n > 3.

a(n) >= n + A006530(A007913(n)) for all nonsquare n. - Peter Kagey, Feb 21 2015

a(n+1) is the smallest number such that the largest prime divisor of a(n) is the highest common factor of a(n) and a(n+1). - Amarnath Murthy, Oct 17 2002

Essentially the same as A036441(n) = a(n+1) and A180107(n) = a(n-1) (n > 1).

The equivalent sequence with A020639 = spf instead of A006530 = gpf begins a(1) = 1, a(2) = 2, and from then on we get all even numbers: a(n) = a(2) + 2*(n-2), n > 1. - M. F. Hasler, Apr 08 2015

From David James Sycamore, Apr 27 2017: (Start)

The sequence contains only one prime; a(2)=2, all other terms (excluding a(1)=1) being composite, since if a(n) for some n > 2 is assumed to be the first prime after 2, then a(n) = a(n-1) + gpf(a(n-1))= m*q+q = q*(m+1) for some integer m > 1 and some prime q. This number is composite; contradiction. Terms after a(3)=4 alternate between even and odd values since each is created by addition of a prime (odd term).

All terms a(n) arise as consecutive multiples of consecutive primes occurring in their natural ascending order, 2,3,5,7.... (A000040). The number of (consecutive) terms which arise as multiples of p(n)= A000040(n) is 1 + p(n+1)- p(n-1), namely n-th term of the sequence: 2,4,5,7,7,7,7,7,11, etc. Example: Number of multiples of 17, the 7th prime, is 1+p(8)-p(6) = 1+19-13 = 7.

For any pair of consecutive primes, p,q (p < q) a(p+q-1) = p*q, the (semiprime) term where multiples of p end and multiples of q start. Example a(7+11-1) = a(17) = 77 = 11*7, the last multiple of 7 and first multiple of 11. Every string of multiples of prime p contains the term p^2, located at a(2*p-1). E.g.: a(3)=4, a(5)=9, a(9)=25. (End)

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