[seqfan] Re: Little puzzle

[Paul Hanna]

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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> Seqfan Mailing list - http://list.seqfan.eu/
>