[seqfan] Re: C(2n,n) - n^2 and C(n^2,n) - n^n
Sean A. Irvine
sairvin at gmail.com
Fri Mar 23 01:31:16 CET 2018
For (ii) I agree with the extra terms found by Hugo and can confirm by
explicit factorization that all of 59, 61, 65, 67, 71, 73, 77, 79, 97 are
definite members (factorizations added to factordb.com). I haven't been
able to complete 83, 85, and 89, but I'm confident they are members of this
sequence.
On 22 March 2018 at 12:21, Hugo Pfoertner <yae9911 at gmail.com> wrote:
> (ii) continues
> 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 97, ... with high probability,
> i.e., if composites of 100+ digits don't have square factors of ~= 30
> digits.
> To be re-checked, many manual copies and pastes involved.
>
> Hugo Pfoertner
>
> On Tue, Mar 20, 2018 at 6:59 PM, юрий герасимов <2stepan at rambler.ru>
> wrote:
>
> > Dear SeqFans, numbers n such that:
> > (i) C(2n,n) - n^2 is prime: 2, 3, 5, 9, 13, 27, 47, 59, 111, 547, 923,
> ...
> > (ii) C(n^2,n) - n^n is squarefree: 2, 3, 5, 7, 13, 17, 19, 23, 29, 31,
> 33,
> > 35,
> > 37, 41, 43, 47, 53,...
> > What tn the next one? Thanks. JSG
> >
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> >
>
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