OFFSET
1,1
COMMENTS
Numbers m with four or more distinct prime factors such that their arithmetic derivative (A003415) can be formed as a carryless (or "carry-free") sum (in the primorial base, A049345) of the respective summands. See the example.
The terms are all squarefree and even (see A380468 and A380478 to find out why). Moreover, they are all multiples of six, because A380459(n) = Product_{d|n} A276086(n/d)^A349394(d) applied to a product of 2*p*q*r, with p, q, r three odd primes > 3 would yield three subproducts which would be multiples of 3 (consider A047247), so the 3-adic valuation of the whole product would be >= 3; hence the second smallest prime factor must be 3. For a similar reason, with terms that are product of four primes, the two remaining prime factors are either both of the form 6m+1 (A002476), or they are both of the form 6m-1 (A007528).
LINKS
EXAMPLE
4686 = 2*3*11*71 and taking subproducts of three primes at time, we obtain 2*3*11 = 66, 2*3*71 = 426, 2*11*71 = 1562, 3*11*71 = 2343. Then A380459(4686) = A276086(66) * A276086(426) * A276086(1562) * A276086(2343) = 1622849599205985150 = 2^1 * 3^2 * 5^2 * 7^6 * 11^9 * 13^1, and because all the exponents are less than the corresponding primes, the product is in A048103.
Considering the primorial base expansions of the same summands (subproducts), we obtain
2100 = A049345(66)
20100 = A049345(426)
73010 = A049345(1562)
101011 = A049345(2343)
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Note how the primorial base digits at the bottom are the exponents in the product A380459(4686) given above, read from the largest to the smallest prime factor
PROG
CROSSREFS
Conjectured to be a subsequence of A046386.
KEYWORD
nonn
AUTHOR
Antti Karttunen, Feb 04 2025
STATUS
approved
