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If n = product{primes p(k)|n} p(k)^b(n,p(k)), where p(k) is the k-th prime that divides n (when these primes are listed from smallest to largest) and each b(n,p(k)) is a positive integer, then the sequence contains the non-prime-powers n such that p(k)^b(n,p(k)) < p(k+1) for all k, 1<=k<= -1 + number of distinct prime divisors of n.
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%I #13 May 16 2017 09:04:53

%S 6,10,14,15,18,20,21,22,26,28,30,33,34,35,38,39,42,44,46,50,51,52,54,

%T 55,57,58,62,65,66,68,69,70,74,75,76,77,78,82,85,86,87,88,91,92,93,94,

%U 95,98,99,100,102,104,105,106,110,111,114,115,116,117,118,119,122,123,124

%N If n = product{primes p(k)|n} p(k)^b(n,p(k)), where p(k) is the k-th prime that divides n (when these primes are listed from smallest to largest) and each b(n,p(k)) is a positive integer, then the sequence contains the non-prime-powers n such that p(k)^b(n,p(k)) < p(k+1) for all k, 1<=k<= -1 + number of distinct prime divisors of n.

%e 2900 is factored as 2^2 * 5^2 * 29^1. Since 2^2 < 5 and 5^2 < 29, then 2900 is in the sequence. On the other hand, 60 is factored as 2^2 * 3^1 * 5^1. Even though 3^1 < 5, 2^2 is not < 3. So 60 is not in the sequence.

%t okQ[n_] := With[{f = FactorInteger[n]}, If[Length[f] == 1, Return[False]]; For[i = 1, i < Length[f], i++, If[f[[i, 1]]^f[[i, 2]] >= f[[i+1, 1]], Return[False]]]; True]; Select[Range[200], okQ] (* _Jean-François Alcover_, May 16 2017, adapted from PARI *)

%o (PARI) isok(n) = {my(f = factor(n)); if (#f~ == 1, return (0)); for (i=1, #f~ - 1, if (f[i, 1]^f[i, 2] >= f[i+1, 1], return (0));); return (1);} \\ _Michel Marcus_, Jan 19 2014

%Y Cf. A057714, A102308.

%K nonn

%O 1,1

%A _Leroy Quet_, Sep 04 2008

%E Extended by _Ray Chandler_, Nov 06 2008