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1
vote
2
answers
501
views
Recurrence for the sum
Let $m\geq 2$ be a fixed integer.
Let
$$f(n):=\begin{cases}
mf\left(\frac{n}{m}\right),&\text{if $n\mod m = 0$;}\\
1,&\text{otherwise}
\end{cases}$$
then if we have
$$a(n):=\begin{cases}
1,&\text{if $ …
0
votes
1
answer
320
views
Generating function for partial sums of the sequence
Let $p$ and $q$ be integers.
Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Then …
2
votes
1
answer
181
views
Pair of recurrence relations with $a(2n+1)=a(2f(n))$
Let $f(n)$ be A053645, distance to largest power of $2$ less than or equal to $n$; write $n$ in binary, change the first digit to zero, and convert back to decimal.
Let $g(n)$ be A007814, the expone …
1
vote
1
answer
294
views
Formula from the recurrence relation
Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Then we have an integer sequence …
3
votes
2
answers
456
views
Subsequence of the cubes
Let $p$ and $q$ be integers.
Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Then …