Re: New arithmetic function

bouayoun nous a récemment amicalement signifié :
Hello,

Hello


I define A new function f by :

I designe p_i a prime, p_1=2, p_2=3,p_3=5, ...

f:NxP --> {0,1} , N natural numbers and P : all prime numbers
f(n,2) = 0 for all n in N*
f(n,p_i) = f(n,p_(i-1)) if p_(i-1) dont divide n
f(n,p_i) = 1-f(n,p_(i-1)) if p_(i-1) divide n

Somme egality is easy but another is hard and I want if this
proposition exist ? is it false or true

1. for all prime , f(p,p) = 0

OK

2. for all prime p,q f(p,q) = 0 if p<=q
f(p,q) = 1 if p>q

I think there is a little error :
f(p_i, p_(i+1)) = 1 - f(p_i, p_i) = 1
So :
for all prime p,q f(p,q) = 0 if p>=q
f(p,q) = 1 if p<q


3. for p,q in P, f(q^n,p) = f(q,p)

OK


4. for a in N* and p in P, f(a,p) = f(rad(a),p) {rad(a) is product
for all prime divisor of a,
example,
rad(12)=rad(3*4)=rad(2*3)=6}

OK




I designe by w+(p) the next prime > p
and by w-(p) the previous prime < p
example w+(7) = 11, w-(7) = 5, w+(14)=15, w-(14)=13
and i note by p# the product of prime , ex: 7# = 2*3*5*7; 3# = 2*3

5. for all a in N* and p in P , f(a+w-(p)#,p) = f(a,p)

OK


6. for all a in N* and p in P and a<= w+(p)#/2 then f(a+w+(p)#/2 ,
p) = f(a,p)

I'm afraid that's wrong
a=2 p=3 w+(p) = 5 w+(p)# = 30 w+(p)#/2 = 15
f(a,p) = f(2, 3) = 1
f(a+w+(p)#/2,p) = f(17, 3) = 0

either I did not understood,
either there are some mistakes
I stop analysing, sorry


[...]


--
Patrick

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