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%I #20 Jan 09 2022 00:54:33
%S 6,10,14,15,21,22,26,30,33,34,35,38,39,42,46,51,55,57,58,60,62,65,66,
%T 69,70,74,77,78,82,84,85,86,87,90,91,93,94,95,102,105,106,110,111,114,
%U 115,118,119,122,123,126,129,130,132,133,134,138,140,141,142,143
%N Numbers n such that the maximal exponent in its prime factorization is less than the number of positive exponents (A051903(n) < A001221(n)).
%C A225230(a(n)) > 1; A050326(a(n)) > 1. - _Reinhard Zumkeller_, May 03 2013
%C Subsequence of A130092. - _Ivan N. Ianakiev_, Sep 17 2019
%D M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 844.
%H Reinhard Zumkeller, <a href="/A212168/b212168.txt">Table of n, a(n) for n = 1..10000</a>
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H Primefan, <a href="http://primefan.tripod.com/500factored.html">The First 2500 Integers Factored</a> (first of 5 pages)
%e 10 = 2^1*5^1 has 2 distinct prime factors, hence 2 positive exponents in its prime factorization (although the 1s are often left implicit). 2 is larger than the maximal exponent in 10's prime factorization, which is 1. Therefore, 10 belongs to the sequence.
%t okQ[n_] := Module[{f = Transpose[FactorInteger[n]][[2]]}, Max[f] < Length[f]]; Select[Range[1000], okQ] (* _T. D. Noe_, May 24 2012 *)
%t Select[Range[200],Max[FactorInteger[#][[All,2]]]<PrimeNu[#]&] (* _Harvey P. Dale_, Nov 21 2018 *)
%o (Haskell)
%o import Data.List (findIndices)
%o a212168 n = a212168_list !! (n-1)
%o a212168_list = map (+ 1) $ findIndices (> 0) a225230_list
%o -- _Reinhard Zumkeller_, May 03 2013
%o (PARI) is(n,f=factor(n))=my(e=f[,2]); #e && vecmax(e)<#e \\ _Charles R Greathouse IV_, Jan 09 2022
%Y Complement of A212165. See also A212164, A212166-A212167.
%Y Subsequence of A188654.
%K nonn
%O 1,1
%A _Matthew Vandermast_, May 22 2012
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