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A090624 If n = Product(pj^ej), i.e., written in its prime factorization, then a(n) = max_j{(pj-1)*ej}.

%I A090624 #28 Apr 29 2023 07:05:30
%S A090624 1,2,2,4,2,6,3,4,4,10,2,12,6,4,4,16,4,18,4,6,10,22,3,8,12,6,6,28,4,30,
%T A090624 5,10,16,6,4,36,18,12,4,40,6,42,10,4,22,46,4,12,8,16,12,52,6,10,6,18,
%U A090624 28,58,4,60,30,6,6,12,10,66,16,22,6,70,4,72,36,8,18,10,12,78,4,8,40,82,6
%N A090624 If n = Product(pj^ej), i.e., written in its prime factorization, then a(n) = max_j{(pj-1)*ej}.
%C A090624 The highest power of k dividing n! (A090622) is close to, but below, n/a(k).
%H A090624 David W. Wilson, <a href="/A090624/b090624.txt">Table of n, a(n) for n = 2..10000</a>
%H A090624 A. M. Oller-Marcen and J. Maria Grau, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Oller/oller3.html">On the Base-b Expansion of the Number of Trailing Zeros of b^k!</a>, J. Int. Seq. 14 (2011) 11.6.8, propos. 3.
%F A090624 a(p) = p-1; a(p^m) = (p-1)*m.
%F A090624 a(b*c) = max(a(b), a(c)) for b and c coprime.
%F A090624 a(n) = lim_{k->inf} k/A090622(k, n) = lim_{k->inf} (k/highest power of k dividing n!). - _David W. Wilson_, Sep 05 2016
%e A090624 72 = 2^3*3^2 so a(72) = max((2-1)*3, (3-1)*2) = max(3,4) = 4.
%p A090624 seq(max(map(t ->(t[1]-1)*t[2], ifactors(n)[2])),n=2..100); # _Robert Israel_, Sep 06 2016
%t A090624 a[n_] := Module[{aux = FactorInteger[n]},Max[Table[aux[[i, 2]]*(aux[[i, 1]] - 1), {i, 1, Length[aux]}]]] (* _José María Grau Ribas_, Feb 15 2010 *)
%o A090624 (PARI) a(n)=my(f=factor(n)); vecmax(vector(#f~,i,(f[i,1]-1)*f[i,2])) \\ _Charles R Greathouse IV_, Sep 07 2016
%o A090624 (Python)
%o A090624 from sympy import factorint
%o A090624 def A090624(n): return max((p-1)*e for p, e in factorint(n).items()) # _Chai Wah Wu_, Apr 28 2023
%K A090624 nonn
%O A090624 2,2
%A A090624 _Henry Bottomley_, Dec 06 2003