OFFSET
1,1
COMMENTS
a(n) is the smallest k such that the minimal infinitary divisor of k! is A050376(n).
Conjecture: a(n) > 0 for all n.
a(28) > 2.5*10^8, if it exists. - Amiram Eldar, Jun 18 2025
a(34) > 2.8*10^10 if it exists. - David A. Corneth, Jun 19 2025
LINKS
EXAMPLE
Let n = 4. A050376(4)=5. For k = 2, 3, 4, 5, 6, we have the following factorizations over distinct terms of A050376: 2! = 2, 3! = 2*3, 4! = 2*3*4, 5! = 2*3*4*5, 6! = 5*9*16. Only the last factorization begins with 5. So a(4) = 6.
From David A. Corneth, Jun 19 2025: (Start)
a(6) = 130. Once we checked that a(6) is > 125 we try 126. The minimal factor of k! into distinct products must be A050376(6) = 8. For 126! we have the 5-adic valuation of 31 so the minimal factor is at most 5.
To get rid of the 5 we try the next candidate > 126 that is a multiple of 5. This is 130. We can just skip 127, 128 and 129 altogether. It turns out this smallest factor for 130! is 8 giving the value for a(6). (End)
PROG
(PARI) \\ See Corneth link
(PARI) \\ See Eldar link
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Vladimir Shevelev, Apr 14 2014
EXTENSIONS
More terms from Peter J. C. Moses, Apr 19 2014
a(19)-a(27) from Amiram Eldar, Jun 18 2025
a(28)-a(35) from David A. Corneth, Jun 18, Jun 21 2025
STATUS
approved