OFFSET
1,2
COMMENTS
Excluding 1, the given terms are products of pairs of numbers from A192580, and are consequently semiprime.
A number n appears in this sequence iff for every prime factor p of n, n/p - 1 is also in the sequence.
For a number n to be expressible as the sum of the partial products of some sequence of primes (i.e., for a number to not appear in this sequence), n must either be prime, or must have a prime factor p such that n = p*(1+x), where x is itself expressible as the sum of partial products of some sequence of primes (i.e., x cannot appear in this sequence). Membership in the sequence can then be shown via a proof by contradiction.
If a(n)+1 is prime, then 2*(a(n)+1) is also in the sequence.
Proof that the sequence is finite and complete: suppose to the contrary there is another term n of the sequence. Then for every prime factor p of n, n/p - 1 must also be in the sequence. Equivalently, for every p, n/p must be in T = {a(k)+1 | 1 <= k <= 16}. It follows that for every p, n/p must divide lcm(T) = 196598730610964320478840. From here there are two cases: (1) If n has only one prime factor p, then n/p | lcm(T). But iterating over all 18 prime power divisors d of lcm(T), for each such that d - 1 is in the sequence, d*p is already in the sequence. (2) If n has two or more prime factors, then the Chinese remainder theorem implies that n | lcm(T). But iterating over all 131072 divisors d of lcm(T), for each such that d/p - 1 is in the sequence for every prime factor p, d is already in the sequence. In both cases, every candidate term n is already present in the sequence. QED.
FORMULA
a(n) = A180337(n) - 1.
EXAMPLE
a(1) = 1 since any sequence of primes will have a sum of partial products greater than 1.
a(2) = 4. Assume to the contrary that 4 = p*(1+x) for some prime p and some x not in this sequence. It immediately follows that p=2, but then x=1, which is in this sequence.
a(3) = 10. Assume to the contrary that 10 = p*(1+x) for some prime p and some x not in this sequence. Then p=2 or p=5, but in the former case, x=4, and in the latter case x=1, and both possible values for x appear in this sequence.
a(5) = 25. Assume to the contrary that 25 = p*(1+x) for some prime p and some x not in this sequence. Then p=5, but then x=4, which appears in this sequence.
Non-examples:
Any sequence composed of a single prime (p) has partial products (p). Therefore no primes appear in this sequence.
The sequence (2, 2) has partial products (2, 4). 2 + 4 = 6, so 6 does not appear in this sequence.
The sequence (2, 3, 2) has partial products (2, 6, 12). 2 + 6 + 12 = 20, so 20 does not appear in this sequence.
PROG
(Python)
from itertools import product
def a():
a192580 = (2, 5, 11, 23, 47)
elems = {1}
for (p, q) in product(a192580, repeat=2):
elems.add(p*q)
return sorted(elems)
(SageMath)
def a():
lcm_T = 196598730610964320478840
result = set()
for n in divisors(lcm_T):
factors = prime_factors(n)
if len(factors) == 1 and n-1 in result:
p = factors[0]
result.add(n*p)
for p in factors:
if n // p - 1 not in result:
break
else:
result.add(n)
return result
CROSSREFS
KEYWORD
nonn,fini,full
AUTHOR
Nick Griffey, Feb 15 2026
STATUS
approved