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

[Hugo Pfoertner]

(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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