cp's OEIS Frontend

Rapidly enters a loop with period 3,5,2,7.

More generally, if a(1) and a(2) are distinct positive numbers with a(1)+a(2) >= 2, the sequence eventually enters the cycle {7,3,5,2} [Back and Caragiu].

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)

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.

More precisely, a(n) = A006530 applied to sum of previous terms.

Inspired by A175723.

Except for initial terms, same as A076272, 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.