A379978 - OEIS

%I #34 Jan 10 2025 20:47:21

%S 1,2,6,12,18,24,50,36,98,48,54,100,242,72,338,196,225,96,578,108,722,

%T 30,441,484,1058,144,250,676,42,392,1682,-1,1922,192,1089,1156,1225,

%U 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

%N 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.

%C a(31) > 26000, if it is not -1. - _Michael S. Branicky_, Jan 08 2025

%C From _Yifan Xie_, Jan 09 2025: (Start)

%C a(31) = -1. Proof:

%C 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.

%C 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.

%C 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)

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

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

%o (Python) # uses code/imports in A066882

%o from itertools import count, islice

%o def agen(limit='float'): # generator of terms

%o     r, n = dict(), 0

%o     for k in count(1):

%o         v = A066882(k)

%o         if v not in r:

%o             r[v] = k

%o             while n in r:

%o                 yield r[n]

%o                 n += 1

%o         if k == limit:

%o             yield from (r[i] if i in r else -1 for i in range(n, max(r)+1))

%o             return

%o print(list(islice(agen(), 31))) # _Michael S. Branicky_, Jan 08 2025

%Y Cf. A066882, A096356.

%K sign,new

%O 0,2

%A _Ilya Gutkovskiy_, Jan 07 2025

%E a(31)-a(53) from _Yifan Xie_, Jan 09 2025

%E a(54)-a(66) from _Alois P. Heinz_, Jan 10 2025