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For prime p in this sequence, b(p) = r*b(p-r) where b(m) = A288814(m), and r = gpf(b(p)) is some prime < q. We can say that prime p_n (n > 2) is of type k if gpf(b(p_n)) = p_(n-k).

Prime gap p-q, and pattern of gaps p-r determines if p is in the sequence or not. Prime p is of type k > 2 only if p-q is one of the even indices of A056240 on which A292081 is defined (12,18,24,28,30,36,...), and if there is a prime r < q < p such that b(p-r) < b(p-q).

Proper subset of A290164.

Terms of A290164 not in this sequence include 5, 11, 61, 191, 431, 541, 1181, 3571, ... corresponding to primes p such that A(4*p) - A(3*p) = A(3*p) - 1, where A=A288814. Examples: A(4*5) - A(3*5) = 51 - 26 = 25; A(4*541) - A(3*541) = 6483 - 3242 = 3241.

These are the primes that require the most effort when searching for the least composite c such that A001414(c) is a given prime, where A001414 is sopfr (sum of prime factors with repetition).

From David James Sycamore, Feb 25 2018: (Start)

Also primes for which A295185 gives a new record.

A006512 gives primes p requiring least effort, since then c=2*(p-2). (End)

Also records of A295185.

Numbers of the form 3^k, 2*3^k, 4*3^k with a(0) = 1 prepended.

If a set of positive numbers has sum n, this is the largest value of their product.

In other words, maximum of products of partitions of n: maximal value of Product k_i for any way of writing n = Sum k_i. To find the answer, take as many of the k_i's as possible to be 3 and then use one or two 2's (see formula lines below).

a(n) is also the maximal size of an Abelian subgroup of the symmetric group S_n. For example, when n = 6, one of the Abelian subgroups with maximal size is the subgroup generated by (123) and (456), which has order 9. [Bercov and Moser] - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 19 2001

Also the maximum number of maximal cliques possible in a graph with n vertices (cf. Capobianco and Molluzzo). - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jul 15 2001 [Corrected by Jim Nastos and Tanya Khovanova, Mar 11 2009]

Every triple of alternate terms {3*k, 3*k+2, 3*k+4} in the sequence forms a geometric progression with first term 3^k and common ratio 2. - Lekraj Beedassy, Mar 28 2002

For n > 4, a(n) is the least multiple m of 3 not divisible by 8 for which omega(m) <= 2 and sopfr(m) = n. - Lekraj Beedassy, Apr 24 2003

Maximal number of divisors that are possible among numbers m such that A080256(m) = n. - Lekraj Beedassy, Oct 13 2003

Or, numbers of the form 2^p*3^q with p <= 2, q >= 0 and 2p + 3q = n. Largest number obtained using only the operations +,* and () on the parts 1 and 2 of any partition of n into these two summands where the former exceeds the latter. - Lekraj Beedassy, Jan 07 2005

a(n) is the largest number of complexity n in the sense of A005520 (A005245). - David W. Wilson, Oct 03 2005

a(n) corresponds also to the ultimate occurrence of n in A001414 and thus stands for the highest number m such that sopfr(m) = n, for n >= 2. - Lekraj Beedassy, Apr 29 2002

a(n) for n >= 1 is a paradigm shift sequence with procedural length p = 0, in the sense of A193455. - Jonathan T. Rowell, Jul 26 2011

a(n) = largest term of n-th row in A212721. - Reinhard Zumkeller, Jun 14 2012

For n >= 2, a(n) is the largest number whose prime divisors (with multiplicity) add to n, whereas the smallest such number (resp. smallest composite number) is A056240(n) (resp. A288814(n)). - David James Sycamore, Nov 23 2017

For n >= 3, a(n+1) = a(n)*(1 + 1/s), where s is the smallest prime factor of a(n). - David James Sycamore, Apr 10 2018

All terms are even. (Cf. formula.)

The definition implies that the sum of factors is the sum over the prime factors with multiplicity, as in A001414. - R. J. Mathar, Nov 28 2008

The sum of factors of a semiprime pq is p+q, which can only be prime if {p, q} = {2, odd prime}. Requiring the sum to be prime then implies that the semiprime is twice the lesser of a twin prime pair. - M. F. Hasler, Apr 07 2015

Subsequence of A288814, each term being of the form A288814(p), where p is the greatest of a pair of twin primes. - David James Sycamore, Aug 29 2017

Sequence is undefined for n=1,2 since no composites exist whose prime divisors sum to 2, 3. For n >= 3, a(n) = A288814(prime(n)) = prime(n-k)*B(prime(n) - prime(n-k)) where B=A056240, and k >= 1 is the "type" of prime(n), indicated as prime(n)~k(g1,g2,...,gk) where gi = prime(n-(i-1)) - prime(n-i); 1 <= i <= k. Thus: 5~1(2), 211~2(12,2), 4327~3(30,8,6) etc. The sequence relates to gaps between odd primes, and in particular to the sequence of k prime gaps below prime(n). The even-indexed terms of B are relevant, as are those of subsequences:

C=A288313, 2,4 plus terms B(n) where n-3 is prime (A298252),

D=A297150, terms B(n) where n-5 is prime and n-3 is composite (A297925) and

E=A298615, terms B(n) where both n-3 and n-5 are composite (A298366).

The above sequences of indices 2m form a partition of the even numbers and the corresponding terms B(2m) form a partition of the even-indexed terms of A056240. The union of D and E is the sequence A292081 = B-C.

Let g(n,t) = prime(n) - prime(n-t), t < n, and h(n,t) = g(n,t) - g(n,1), 1 < t < n. If g1=g(n,1) is a term in A298252 (g1-3 is prime), then B(g1) is a term in C, so k=1. If g1 belongs to A297925 or A298366 then B(g1) is a term in D or E and the value of k depends on subsequent gaps below prime(n), within a range dependent on g1.

Let range R1(g1) = u - g(n,1) where u is the index in B of the greatest term in C such that C(u) < B(g1). Let range R2(g1) = v-g(n,1) where v is the index in B of the greatest term in D such that D(v) <= B(g1). For all n, R2 < R1, and if g1 is a term in D then R2(g1)=0. Examples: R1(12)=2, R2(12)=0, R1(30)=26, R2(30)=6.

k >= 1 is the smallest integer such that B(g(n,k)) <= B(g(n,t)) for all t satisfying g1 <= g(n,t) <= g1 + R1(g1). For g1-3 prime, k=1. If g1-3 is composite, let z be least integer > 1 such that g(n,z)-3 is prime, and let w be least integer >= 1 such that g(n,w)-5 is prime. Then z "complies" if h(n,z) <= R1, and w "complies" if h(n,w) <= R2. If g1-5 is prime then R2=w=0 and only z is relevant.

B(g1) must belong to C,D or E. If in C (g1-3 is prime) then k=1. If in D (g1-5 is prime), k=z if z complies, otherwise k=1. If B(g1) is in E and z complies but not w then k=z, or if w complies but not z then k=w. If B(g1) is in E and z,w both comply then k=z if 3*(g(n,z)-3) < 5*(g(n,w)-5), otherwise k=w. If neither z nor w comply, then k=1.

Conjecture: For all n >= 3, a(n) >= A288189(n).

A010051(2*a(n) - 3) * A010051(3*a(n) - 2) = 1;

A259758(n) = (2*a(n) - 3) * (3*a(n) - 2).

Except for a(1)=3 this is the same sequence as primes p such that A288814(3*p) - A288814(2*p) = 5. - David James Sycamore, Jul 22 2017

Furthermore, (A288814(3*p)*A288814(2*p))/6 belongs to A259758. - David James Sycamore, Jul 23 2017

Least semiprime whose sum of prime factors equals 2*n.

Assuming Goldbach's conjecture, a(n) exists for all n >= 2. - David James Sycamore, Jan 08 2019