[seqfan] Re: Can we always reach a power of 2 with this formula?
Ami Eldar
amiram.eldar at gmail.com
Sun Dec 26 08:55:45 CET 2021
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
>
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> Seqfan Mailing list - http://list.seqfan.eu/
>
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