# [seqfan] Re: Fwd: Formula for A025672

Sean A. Irvine sairvin at xtra.co.nz
Mon Feb 20 03:33:04 CET 2012

```Yes, it is wrong.  The first disagreement is at term 110; and it gets
progressively worse from there.

Here are the first few disagreements:

110 mirceau=0 truth=10
111 mirceau=10 truth=0
121 mirceau=0 truth=10
122 mirceau=10 truth=0
131 mirceau=1 truth=11
132 mirceau=0 truth=1
133 mirceau=11 truth=10
134 mirceau=10 truth=0
143 mirceau=1 truth=11
144 mirceau=0 truth=1
145 mirceau=11 truth=10
146 mirceau=10 truth=0

franktaw at netscape.net wrote:
> I am certain it is wrong. The formula (correcting the final k to n) is
> for the sequence which concatenates
> <n-1,n-2,n-3,...,0,n-1,n-2,n-3,...,0> for each n. A025672 depends on
> log(3)/log(8), which is an irrational number, so it does not have this
> form. I don't where they diverge, but I would guess it's not much past
> the published sequence.
>
>
> -----Original Message-----
> From: David Wilson <davidwwilson at comcast.net>
>
> The sender already A025672 the sequence with the following formula.
>
> My intuition says the formula is wrong.
>
> Any takers?
>
> -------- Original Message --------
> Subject:     Formula for A025672
>
> Dear David W. Wilson,
>
> I am glad to say I found a formula for the sequence A025672 .
>
> A025672(n)  =  floor((1/4)*ceil(2*sqrt(n))^2)-k
>
> I hope it is useful.
>
> Sincerely,
>
> Mircea Merca,
>
> Department of Informatics,
>
> Constantin Istrati Technical College,
>
> 105600 Campina, Romania
>
>
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```