[seqfan] Re: Can we always reach a power of 2 with this formula?
Fred Lunnon
fred.lunnon at gmail.com
Sun Dec 26 12:11:40 CET 2021
<< Can we prove that we can always find a value for a(n)? >>
It looks to be an elementary consequence of Fermat little theorem, Hensel's
lemma,
and chinese remainder theorem, that more generally for every natural u,v
there exist x,y
such that x*u = (2^y - 1)2^v . WFL
On Sun, Dec 26, 2021 at 9:49 AM Ami Eldar <amiram.eldar at gmail.com> wrote:
> Hello Ali,
>
> You have an error in a(4). It is 12 and not 8 (2^4 + 8*20 = 176 and not
> 256).
>
> The first 30 terms are
> 1, 2, 2, 12, 16, 96, 16, 224, 23296, 9761280, 15872, 107520, 184320,
> 319488, 2048, 61440, 7186350080, 200933376, 94812623732736, 10223616,
> 4874025566208, 10759916740583034417814216835787128832, 63739789312,
> 29320282112, 59516348042566359318528, 6569353302507520, 46539997184,
> 6394454259279344370111448350720, 165649640849408, 1210643906560
>
> Best,
> Amiram
>
>
> On Sun, Dec 26, 2021 at 9:40 AM Ali Sada via SeqFan <seqfan at list.seqfan.eu
> >
> wrote:
>
> > Hi everyone,
> >
> > a(n) is the least positive integer such that 2^n + a(n)*A002378(n) equals
> > to a power of 2.
> >
> > 2^1 + 1*2 = 4, a(1) = 1
> > 2^2 + 2*6 = 16, a(2) = 2
> > 2^3 + 2*12 = 32, a(3) = 2
> > 2^4 + 8*20 = 256, a(4) = 8
> > 2^5 + 16*30 = 512, a(5) = 16
> > 2^6 + 96*42 = 4096, a(6) = 96
> > 2^7 + 16*56 = 1024, a(7) = 16
> >
> > Can we prove that we can always find a value for a(n)?
> > I would like to add this sequence to the OEIS and I would really
> > appreciate it if someone could calculate the necessary terms if possible.
> >
> > Best,
> >
> > Ali
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>
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