[seqfan] Re: Little puzzle
Luca Petrone
luca.petrone at libero.it
Thu Apr 12 02:17:21 CEST 2018
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)
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
> > >
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> Seqfan Mailing list - http://list.seqfan.eu/
>
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