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A100574 If n = product{p|n, p=prime} p^b(p,n), where each b(p,n) is a positive integer and the product is over distinct prime divisors of n, a(n) = difference between the maximum p^b(p,n) and minimum p^b(p,n).

%I A100574 #11 Aug 06 2018 22:39:28
%S A100574 0,0,0,0,0,1,0,0,0,3,0,1,0,5,2,0,0,7,0,1,4,9,0,5,0,11,0,3,0,3,0,0,8,
%T A100574 15,2,5,0,17,10,3,0,5,0,7,4,21,0,13,0,23,14,9,0,25,6,1,16,27,0,2,0,29,
%U A100574 2,0,8,9,0,13,20,5,0,1,0,35,22,15,4,11,0,11,0,39,0,4,12,41,26,3,0,7,6,19,28
%N A100574 If n = product{p|n, p=prime} p^b(p,n), where each b(p,n) is a positive integer and the product is over distinct prime divisors of n, a(n) = difference between the maximum p^b(p,n) and minimum p^b(p,n).
%C A100574 a(n)=0 iff n a prime or a power of a prime. - _Robert G. Wilson v_, Jan 10 2005
%H A100574 Antti Karttunen, <a href="/A100574/b100574.txt">Table of n, a(n) for n = 1..65537</a>
%F A100574 a(n) = A034699(n) - A034684(n). - _Antti Karttunen_, Aug 06 2018
%e A100574 For 24 = 2^3 *3, 2^3 and 3 are separated by 5, so a(30) = 5.
%t A100574 pf[n_] := Block[{pb = Flatten[ Table[ #[[1]]^#[[2]], {1}] & /@ FactorInteger[n]]}, Max[pb] - Min[pb]]; Table[ pf[n], {n, 2, 100}] (* _Robert G. Wilson v_, Jan 10 2005 *)
%o A100574 (PARI) A100574(n) = if(1==n,0,my(f=factor(n), v = vector(#f[, 1], i, f[i, 1]^f[i, 2])); vecmax(v)-vecmin(v)); \\ _Antti Karttunen_, Aug 06 2018
%Y A100574 Cf. A034684, A034699.
%K A100574 nonn
%O A100574 1,10
%A A100574 _Leroy Quet_, Jan 02 2005
%E A100574 More terms from _Robert G. Wilson v_, Jan 10 2005