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Dougy · 2005-09-26, 06:48

Just to clarify. Let $N$ be any natural number. It is clear that the number of digits of $N$ , $D(N)$ , satisfies $D(N)= \left\lfloor log_{10}(N)+1\right\rfloor$ .

For Riesel-type numbers $N=k \cdot 2^n-1$ then $D(N)=\left\lfloor log_{10}(N)+1\right\rfloor = \left\lfloor log_{10}(k \cdot 2^n-1)+1\right\rfloor=\left\lfloor log_{10}(k \cdot 2^n)+1\right\rfloor=\left\lfloor log_{10}(k)+ log_{10}(2^n)+1\right\rfloor=\left\lfloor log_{10}(k)+ n \cdot log_{10}(2)+1\right\rfloor$ , using log laws, except when $k=5^n$ (i.e. $N+1=k \cdot 2^n = 10^n$ ) when we've overcalculated by one digit, and for this special case $D(N)=\left\lfloor log_{10}(k)+ n \cdot log_{10}(2)\right\rfloor$ .

For Proth-type numbers, $N=k \cdot 2^n+1$ and $D(N)= \left\lfloor log_{10}(N)+1\right\rfloor=\left\lfloor log_{10}(k \cdot 2^n+1)+1\right\rfloor=\left\lfloor log_{10}(k \cdot 2^n)+1\right\rfloor=\left\lfloor log_{10}(k)+ log_{10}(2^n)+1\right\rfloor=\left\lfloor log_{10}(k)+ n \cdot log_{10}(2)+1\right\rfloor$ . There are no special cases here since a special case would require $N$ to be written $10000...000$ in decimal notation, which is impossible since $N$ is odd.

2005-09-26, 06:48	#8
Dougy Aug 2004 Melbourne, Australia 152₁₀ Posts	Just to clarify. Let $N$ be any natural number. It is clear that the number of digits of $N$ , $D(N)$ , satisfies $D(N)= \left\lfloor log_{10}(N)+1\right\rfloor$ . For Riesel-type numbers $N=k \cdot 2^n-1$ then $D(N)=\left\lfloor log_{10}(N)+1\right\rfloor = \left\lfloor log_{10}(k \cdot 2^n-1)+1\right\rfloor=\left\lfloor log_{10}(k \cdot 2^n)+1\right\rfloor=\left\lfloor log_{10}(k)+ log_{10}(2^n)+1\right\rfloor=\left\lfloor log_{10}(k)+ n \cdot log_{10}(2)+1\right\rfloor$ , using log laws, except when $k=5^n$ (i.e. $N+1=k \cdot 2^n = 10^n$ ) when we've overcalculated by one digit, and for this special case $D(N)=\left\lfloor log_{10}(k)+ n \cdot log_{10}(2)\right\rfloor$ . For Proth-type numbers, $N=k \cdot 2^n+1$ and $D(N)= \left\lfloor log_{10}(N)+1\right\rfloor=\left\lfloor log_{10}(k \cdot 2^n+1)+1\right\rfloor=\left\lfloor log_{10}(k \cdot 2^n)+1\right\rfloor=\left\lfloor log_{10}(k)+ log_{10}(2^n)+1\right\rfloor=\left\lfloor log_{10}(k)+ n \cdot log_{10}(2)+1\right\rfloor$ . There are no special cases here since a special case would require $N$ to be written $10000...000$ in decimal notation, which is impossible since $N$ is odd.