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P(n!+1)-P(2^n+1) where P(x) is the largest prime factor in x.
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%I #5 Mar 30 2012 18:38:58

%S -1,-2,4,-12,0,90,28,404,250,329850,39916118,2834088,75021616,

%T 3790360374,46271010,993974,956666,123610842,1713311273189068,

%U 117876621366,2703875810364,93799610095767534,148139754734068388,765041185860961083618,38681321803817920155550

%N P(n!+1)-P(2^n+1) where P(x) is the largest prime factor in x.

%C Is always true that a(n)>0 for n>5? More generally, if m is an integer >2, is there always an integer f(m) such that P(n!+1)>P(m^n+1) for n>f(m) (it seems that f(2)=5, f(3)=7, f(4)=17, ...)

%K easy,sign

%O 1,2

%A _Benoit Cloitre_, May 12 2002