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Max Alekseyev
Max Alekseyev
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ThisThe formula $c(n)=a(n+1)$ is pretty much straightforward, noticing that $$\lfloor \log_{i+1}(n-i)\rfloor - \lfloor\log_{i+1}(n-1-i)\rfloor=1\quad\text{iff}\quad n-i=(i+1)^m\text{ for some }m.$$ The latter condition means that $n+1=k+k^m$ with $k:=i+1$.

This is pretty much straightforward, noticing that $$\lfloor \log_{i+1}(n-i)\rfloor - \lfloor\log_{i+1}(n-1-i)\rfloor=1\quad\text{iff}\quad n-i=(i+1)^m\text{ for some }m.$$ The latter condition means that $n+1=k+k^m$ with $k:=i+1$.

The formula $c(n)=a(n+1)$ is pretty much straightforward, noticing that $$\lfloor \log_{i+1}(n-i)\rfloor - \lfloor\log_{i+1}(n-1-i)\rfloor=1\quad\text{iff}\quad n-i=(i+1)^m\text{ for some }m.$$ The latter condition means that $n+1=k+k^m$ with $k:=i+1$.

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Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152

This is pretty much straightforward, noticing that $$\lfloor \log_{i+1}(n-i)\rfloor - \lfloor\log_{i+1}(n-1-i)\rfloor=1\quad\text{iff}\quad n-i=(i+1)^m\text{ for some }m.$$ The latter condition means that $n+1=k+k^m$ with $k:=i+1$.