[seqfan] Re: C(2n,n) - n^2 and C(n^2,n) - n^n

[Hugo Pfoertner]

Both to be submitted? Yuri, Harvey, Sean ...

On Fri, Mar 23, 2018 at 1:31 AM, Sean A. Irvine <sairvin at gmail.com> wrote:

> 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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>
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