Is this identity true?
Look at attachment
M(50) = mu(27)+mu(29)+mu(31)+....+mu(49)
= 0+(-1)+(-1)+......+0= -3
mu(n) is Mobius function
It is true!
People checked it in other fora...
So thanx for reading the post.
The traditional definition: M(n) = mu(1)+mu(2)+...+mu(n)
for n=8 we have M(8) = 1+(-1)+(-1)+0+(-1)+1+(-1)+0 = -2
Your formula Gh(8) = mu(4+2)+mu(4+4) = mu(6)+mu(8) = 1
in contradiction to the traditional formula
n=8=2*4 an 4 is not odd
Read the condition : o must be odd >=3 then you can compute M(2*o)
My formula holds. Someone in another forum just proved it.
I have a proof but it is little bit long.
Thank you for your comment
With n=2*o, (o odd) it's OK
For my investigation I used ARIBAS (Windows version) and I programmed a function 'SmallMoebiusMu(n)' (small because n must not exceed 2**32) and with this function, I compared my function 'SmallMertensNumber(n)' (traditional definition) to the function 'Gaussianheart(n)' (your formulation) and for n=2,6,10,14,18,...,402 I found equal results.
SmallMoeniusMu uses the built-in ARIBAS function 'factor16' and 'prime32test'.
Regards from Germany
The formula is correct!
Good for me!
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