A379978 a(n) is the smallest positive integer which can be represented as the sum of its prime divisors in exactly n ways, or -1 if no such integer exists.

1, 2, 6, 12, 18, 24, 50, 36, 98, 48, 54, 100, 242, 72, 338, 196, 225, 96, 578, 108, 722, 30, 441, 484, 1058, 144, 250, 676, 42, 392, 1682, -1, 1922, 192, 1089, 1156, 1225, 216, 2738, 1444, 1521, 400, 66, 70, 3698, 968, 675, 2116, 4418, 78, 686, 500, 2601, 1352, 5618, 324, 3025, 784, 3249, 3364, 6962, 105, 7442, 102, 1323, 110, 4225

Offset: 0

Author

Ilya Gutkovskiy, Jan 07 2025

Keywords

• sign

Comments

a(31) > 26000, if it is not -1. - Michael S. Branicky, Jan 08 2025

From Yifan Xie, Jan 09 2025: (Start)

a(31) = -1. Proof:

Lemma: n can be partitioned into a and b (gcd(a, b)=1) if n>ab-a-b. Proof: Using Bezout's theorem we can get n = xa+yb for integers x and y. Substitute x'=x-kb for x and y'=y+ka for y such that 0 <= x' <= b-1, then y'>-1, so y'>=0, a valid partition.

Suppose that a(31)=n has 3 distinct prime divisors ppq-p-q. so c<(n+p+q-pq)/r, hence there are at least (n+p+q-pq)/r choices for c, but there are at most 31 partitions, thus 31r >= n+p-q(p-1) >= n-r*(p-1) >= r*(pq+1-p), p*(q-1) <= 30, and r <= 30 since there are r+1 partitions of pqr into p and q. Enumerate all possibilities of p, q, r and only (p,q,r) = (2,3,5) and (2,3,7) give no more than 31 partitions. But in these cases, if n=pqr, there are fewer than 31 partitions; if n >= 2pqr, there are more than 31 partitions.

If a(31) = n has exactly 2 prime divisors p, q, it's easy to see that n has n/(p*q) + 1 partitions into p and q. therefore n = 30*p*q, a contradiction. If a(31) = n is prime, n has only 1 partition. (End)

Examples

			a(3) = 12: 12 = 2 + 2 + 2 + 2 + 2 + 2 = 2 + 2 + 2 + 3 + 3 = 3 + 3 + 3 + 3.
		

Crossrefs

Cf. A066882, A096356.

Programs

Python

# uses code/imports in A066882
from itertools import count, islice
def agen(limit='float'): # generator of terms
    r, n = dict(), 0
    for k in count(1):
        v = A066882(k)
        if v not in r:
            r[v] = k
            while n in r:
                yield r[n]
                n += 1
        if k == limit:
            yield from (r[i] if i in r else -1 for i in range(n, max(r)+1))
            return
print(list(islice(agen(), 31))) # Michael S. Branicky, Jan 08 2025

Formula

If a(n) > 0, A066882(a(n)) = n.

Extensions

a(31)-a(53) from Yifan Xie, Jan 09 2025

a(54)-a(66) from Alois P. Heinz, Jan 10 2025