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Adam

My ol mate strikes again, without the delta:

$$d(x,p,j)=\Bigl\lfloor{x\,{p}^{\,j- {\lfloor\ln_p(x)\rfloor} -1}} \Bigr\rfloor-p\Bigl\lfloor{x\,{p}^{\,j- {\lfloor\ln_p(x)\rfloor} -2}} \Bigr\rfloor$$

$$x=\sum_{j=1}^{\infty}d(x,p,j)\,p^{\,\lfloor \ln_p(x)\rfloor+1-j}$$

$$\lfloor x\rfloor=\sum_{j=1}^{\lfloor\ln_p(x)\rfloor+1}d(x,p,j)\,p^{\,\lfloor\ln_p(x)\rfloor+1-j}$$

$${{x}}=\sum_{j=\lfloor \ln_p(x)\rfloor+2}^{\infty}d(x,p,j)\,p^{\,\lfloor \ln_p(x)\rfloor+1-j}$$ $$\forall x \in \mathbb R^+$$

The following is of interest in connecting the above in a useful sense to the prime number problem i began work on a few months ago:

$$\vartheta_{n,m,k}=\Biggl\lfloor \Biggl(n^{\frac{m}{k}} -\lfloor{\lfloor n^{\frac{m}{k}} \rfloor}^{\frac{k}{m}-1}\rfloor\gcd\Bigl(\lfloor{\lfloor n^{\frac{m}{k}} \rfloor}^{\frac{k}{m}-1}\rfloor,\Bigl\lfloor\Bigl\lfloor \frac{p_n^{\frac{m}{k}}}{n^{\frac{m}{k}}} \Bigr\rfloor^{\frac{k}{m}-1}\Bigr\rfloor\Bigr)\Biggr)^{\frac{k}{m}}\Biggr\rfloor $$

$$m \geq k \Rightarrow n-\vartheta_{n,m,k} \in {{0,1}}$$

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