[seqfan] Re: Little puzzle

[Luca Petrone]

Dear Paul,

actually, according to David's definition sum is from 1 to infinity, not from 0, so

(1) a(n) = suminf(k=1,floor(n/2^k+1/2))

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,...]

(which is consistent with sample for f(11) = (6, 3, 1, 1, 0, 0, 0, ...) which sum is 11)

so a(n) = n

(2) b(n) = suminf(k=1, 2^k*floor(n/2^k +1/2)^2)

b = [2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, ...]

so b(n) = n*(n+1)

Luca

> 
>     Il 11 aprile 2018 alle 22.32 Paul Hanna <pauldhanna.math at gmail.com> ha scritto:
> 
>     Hi David,
>     Using PARI, as I understand the puzzle, I would write
> 
>     (1) For SUM(k = 1..inf, f(k)) :
> 
>     a(n) = suminf(k=0, floor(n/2^k + 1/2))
> 
>     a = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, ...]
> 
>     so it appears that a(n) = 2*n .
> 
>     (2) For SUM(k = 1..inf, 2^k f(k)^2) :
> 
>     b(n) = suminf(k=0, 2^k*floor(n/2^k +1/2)^2)
> 
>     b = [3, 10, 21, 36, 55, 78, 105, 136, 171, 210, ...]
> 
>     and it appears that these are triangular numbers where
> 
>     b(n) = n*(2*n+1) .
> 
>     Did I get it right?
> 
>     Nice puzzle.
>     Paul
> 
>     On Tue, Apr 10, 2018 at 5:57 PM, David Wilson <davidwwilson at comcast.net>
>     wrote:
> 
>         > > 
> >         Let r(k) = [k + 1/2] = k rounded to the nearest integer.
> > 
> >         For integer n >= 0, define the sequence
> > 
> >         S(k) = { r(k/2^n) : n >= 1 }
> > 
> >         So, for example, we have
> > 
> >         f(11) = (r(11/2), r(11/4), r(11/8), ...) = (6, 3, 1, 1, 0, 0, 0, ...)
> > 
> >         What are
> > 
> >         SUM(k = 1..inf, f(k))
> >         SUM(k = 1..inf, 2^k f(k)^2)
> > 
> >         --
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> > 
> >     > 
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