Does this sum go infinity?












6












$begingroup$


Consider $F(x)$ that maps from $Bbb N$ to ${pm 1}$, such that if $x$ is odd, then $F(x)$ = $$(-1)^{(frac{x-1}{2})}$$, and if $x$ is even, then $F(x)=F(y)$, where $y$ is the odd number obtained after dividing $x$ by $2$ until it is odd.



Does $S_n = sum_{p=1}^n F(p)$ have an explicit formula?
And if $n$ tends to $infty$ does the sum alternates or will it lie between an interval or does it tend to $pm infty$ ?






sequences-and-series algebra-precalculus







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    $begingroup$
    @Max thanks for the edit....this was my first question on mathstackexchange!
    $endgroup$
    – Hari Krishna P
    4 hours ago
















6












$begingroup$


Consider $F(x)$ that maps from $Bbb N$ to ${pm 1}$, such that if $x$ is odd, then $F(x)$ = $$(-1)^{(frac{x-1}{2})}$$, and if $x$ is even, then $F(x)=F(y)$, where $y$ is the odd number obtained after dividing $x$ by $2$ until it is odd.



Does $S_n = sum_{p=1}^n F(p)$ have an explicit formula?
And if $n$ tends to $infty$ does the sum alternates or will it lie between an interval or does it tend to $pm infty$ ?






sequences-and-series algebra-precalculus







share|cite|improve this question









New contributor




Hari Krishna P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    @Max thanks for the edit....this was my first question on mathstackexchange!
    $endgroup$
    – Hari Krishna P
    4 hours ago














6












6








6


3



$begingroup$


Consider $F(x)$ that maps from $Bbb N$ to ${pm 1}$, such that if $x$ is odd, then $F(x)$ = $$(-1)^{(frac{x-1}{2})}$$, and if $x$ is even, then $F(x)=F(y)$, where $y$ is the odd number obtained after dividing $x$ by $2$ until it is odd.



Does $S_n = sum_{p=1}^n F(p)$ have an explicit formula?
And if $n$ tends to $infty$ does the sum alternates or will it lie between an interval or does it tend to $pm infty$ ?






sequences-and-series algebra-precalculus







share|cite|improve this question









New contributor




Hari Krishna P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Consider $F(x)$ that maps from $Bbb N$ to ${pm 1}$, such that if $x$ is odd, then $F(x)$ = $$(-1)^{(frac{x-1}{2})}$$, and if $x$ is even, then $F(x)=F(y)$, where $y$ is the odd number obtained after dividing $x$ by $2$ until it is odd.



Does $S_n = sum_{p=1}^n F(p)$ have an explicit formula?
And if $n$ tends to $infty$ does the sum alternates or will it lie between an interval or does it tend to $pm infty$ ?






sequences-and-series algebra-precalculus




sequences-and-series algebra-precalculus






share|cite|improve this question









New contributor




Hari Krishna P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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edited 4 hours ago







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asked 4 hours ago









Hari Krishna PHari Krishna P

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314




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  • 1




    $begingroup$
    @Max thanks for the edit....this was my first question on mathstackexchange!
    $endgroup$
    – Hari Krishna P
    4 hours ago














  • 1




    $begingroup$
    @Max thanks for the edit....this was my first question on mathstackexchange!
    $endgroup$
    – Hari Krishna P
    4 hours ago








1




1




$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
4 hours ago




$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
4 hours ago










1 Answer
1






active

oldest

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5












$begingroup$

Let $T(n) = sum_{ktext{ is odd, } kle n}F(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.



Then, $T(lfloor n/2rfloor) = sum_{ktext{ is odd, } kle lfloor n/2rfloor}F(k)=sum_{ktext{ is odd, } 2kle n}F(2k)$, $T(lfloor n/4rfloor) = sum_{ktext{ is odd, } 4kle n}F(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_{m=0}^{lfloor log_2 nrfloor}T(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.



Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^{m-1}$ place, is always there) or as high as $m$ for $n=lfloor 2^{m+1}/3rfloor$.



So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.






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  • $begingroup$
    Awesome! Thank you!!
    $endgroup$
    – Hari Krishna P
    2 hours ago











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1 Answer
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5












$begingroup$

Let $T(n) = sum_{ktext{ is odd, } kle n}F(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.



Then, $T(lfloor n/2rfloor) = sum_{ktext{ is odd, } kle lfloor n/2rfloor}F(k)=sum_{ktext{ is odd, } 2kle n}F(2k)$, $T(lfloor n/4rfloor) = sum_{ktext{ is odd, } 4kle n}F(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_{m=0}^{lfloor log_2 nrfloor}T(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.



Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^{m-1}$ place, is always there) or as high as $m$ for $n=lfloor 2^{m+1}/3rfloor$.



So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Awesome! Thank you!!
    $endgroup$
    – Hari Krishna P
    2 hours ago
















5












$begingroup$

Let $T(n) = sum_{ktext{ is odd, } kle n}F(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.



Then, $T(lfloor n/2rfloor) = sum_{ktext{ is odd, } kle lfloor n/2rfloor}F(k)=sum_{ktext{ is odd, } 2kle n}F(2k)$, $T(lfloor n/4rfloor) = sum_{ktext{ is odd, } 4kle n}F(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_{m=0}^{lfloor log_2 nrfloor}T(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.



Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^{m-1}$ place, is always there) or as high as $m$ for $n=lfloor 2^{m+1}/3rfloor$.



So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Awesome! Thank you!!
    $endgroup$
    – Hari Krishna P
    2 hours ago














5












5








5





$begingroup$

Let $T(n) = sum_{ktext{ is odd, } kle n}F(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.



Then, $T(lfloor n/2rfloor) = sum_{ktext{ is odd, } kle lfloor n/2rfloor}F(k)=sum_{ktext{ is odd, } 2kle n}F(2k)$, $T(lfloor n/4rfloor) = sum_{ktext{ is odd, } 4kle n}F(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_{m=0}^{lfloor log_2 nrfloor}T(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.



Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^{m-1}$ place, is always there) or as high as $m$ for $n=lfloor 2^{m+1}/3rfloor$.



So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.






share|cite|improve this answer









$endgroup$



Let $T(n) = sum_{ktext{ is odd, } kle n}F(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.



Then, $T(lfloor n/2rfloor) = sum_{ktext{ is odd, } kle lfloor n/2rfloor}F(k)=sum_{ktext{ is odd, } 2kle n}F(2k)$, $T(lfloor n/4rfloor) = sum_{ktext{ is odd, } 4kle n}F(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_{m=0}^{lfloor log_2 nrfloor}T(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.



Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^{m-1}$ place, is always there) or as high as $m$ for $n=lfloor 2^{m+1}/3rfloor$.



So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.







share|cite|improve this answer












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answered 3 hours ago









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    $endgroup$
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  • $begingroup$
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$begingroup$
Awesome! Thank you!!
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$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
2 hours ago










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