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Let b(n) = number of prime factors (with multiplicity) of concatenation of numbers from 1 to n; sequence gives smallest number m with b(m) = n.
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%I #25 Oct 01 2023 07:34:35

%S 1,3,2,9,8,11,28,6,26,20

%N Let b(n) = number of prime factors (with multiplicity) of concatenation of numbers from 1 to n; sequence gives smallest number m with b(m) = n.

%C a(11) <= 148, a(12) = 104, a(13) = 100, a(14) = 54, a(15) <= 184. - _Michael S. Branicky_, Dec 14 2022 using factorizations at the De Geest link

%H Patrick De Geest, <a href="http://www.worldofnumbers.com/factorlist.htm">Normal Smarandache Concatenated Numbers, Prime factors from 1 up to n</a>

%H Micha Fleuren, <a href="http://www.gallup.unm.edu/~smarandache/michafleuren.htm">Factors and primes of Smarandache sequences</a>.

%H Micha Fleuren, <a href="http://www.gallup.unm.edu/~smarandache/micha.txt">Smarandache Factors and Reverse factors</a>

%H Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_008.htm">Puzzle 8. Primes by Listing</a>, The Prime Puzzles & Problems connection.

%t Join[{1},Table[i=1; While[PrimeOmega[FromDigits[Flatten[IntegerDigits[Range[i]]]]]!=n,i++]; i,{n,2,10}]] (* _Jayanta Basu_, May 30 2013 *)

%Y Cf. A007908, A050678, A046460.

%K nonn,base,hard,more

%O 1,2

%A _Patrick De Geest_, Aug 15 1999