## The ordinary generating function for the Mobius function

$\sum \limits_{n=1}^{\infty} \mu(n)x^n = x - \frac{x^{2}}{1-x} + \sum \limits_{a=2}^{\infty} \frac{x^{2a}}{1-x^{a}} - \sum \limits_{b=2}^{\infty} \sum \limits_{a=2}^{\infty} \frac{x^{2ab}}{1-x^{ab}} + \sum \limits_{c=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{a=2}^{\infty} \frac{x^{2abc}}{1-x^{abc}} - \sum \limits_{d=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{a=2}^{\infty} \frac{x^{2abcd}}{1-x^{abcd}} + ...$

This expression above is basically what I posted to the seqfan mailing list:
seqfan-2011-March-5

And here:
seqfan-2011-March-8

And here:
Wikipedia Möbius function

Wikipedia, Möbius function, difference between revisions 1

Wikipedia, Möbius function, difference between revisions 2

Using the six first terms (on the right hand side) I get 14 accurate decimals, of 14 possible in a spreadsheet, in the interval -0.5 to +0.5. The technique for inverting triangular matrices as a binomial series, which is the basis for this formula, was learned together with Gary W. Adamson.

This expression above simplifies to:

$\sum \limits_{n=1}^{\infty} \mu(n)x^n = x - \sum \limits_{a=2}^{\infty} x^{a} + \sum \limits_{b=2}^{\infty} \sum \limits_{a=2}^{\infty} x^{ab} - \sum \limits_{c=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{a=2}^{\infty} x^{abc} + \sum \limits_{d=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{a=2}^{\infty} x^{abcd} - ...$

mats.granvik(AT)abo.fi

Keywords: Mobius function, power series, binomial series.