п(n)≈n(1-1/2)(1-1/3)(1-1/5)(1-1/7)...(1-1/pп(n1/2))+п(n1/2)-1.
п(n)>=[n(1-1/2)(1-1/3)(1-1/5)(1-1/7)...(1-1/pп(n1/2))]+п(n1/2)-1.
if n€N, f(n)€N then f(n)!=f(1)f(2)f(3)...f(n) is generalization of n!=1*2*3*4*...*n.
n!=n!; (2n)!=2*4*6*...*(2n); (2n-1)!=1*3*5*...*(2n-1); (3n+1)!=4*7*10*...*(3n+1).
f(n)=pn, pn is prime number, p1=2, p2=3, p3=5, p4=7, p5=11,...
п(n)=SUMp<=n1.
pk!=p1p2p3...pk=2*3*5*7*11*...*pk;
(pk-1)!=(p1-1)(p2-1)...(pk-1)=1*2*4*6*10*...*(pk-1).
п(n)≈n(pk-1)!/pk! + п(n1/2)-1, k= п(n1/2).
п(n)≈n([n1/2]!)-1ф([n1/2]!)+ п(n1/2)-1.
п(n)≈SUMk=1ф([n1/2^k]!)n1/2^(k-1)/[n1/2^k]!.
ф(n) is euler's function.
Emzari Papava
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