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

0,2

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 p<q<r, and a valid partition is n=ap+bq+cr. Using the lemma, ap+bq has at least one partition if n-cr>pq-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)

Links

Table of n, a(n) for n=0..66.

Formula

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

Example

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

Prog

(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

Crossrefs

Cf. A066882, A096356.

Sequence in context: A348774 A265843 A276829 * A032371 A162802 A330320

Adjacent sequences: A379975 A379976 A379977 * A379979 A379980 A379981

Keyword

sign,new

Author

Ilya Gutkovskiy, Jan 07 2025

Extensions

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

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

Status

approved