Sequences based on algorithms

[Max Alekseyev]

On Nov 14, 2007 11:14 PM, Jonathan Post <jvospost3 at gmail.com> wrote:

> I just submitted a related seq.  Whether the pattern I suggest for the
> first 5 values (as subset of A132435) continues I leave to others,
> perhaps through programming it, though I should have clicked "more" to
> invite that.
>
> %I A000001
> %S A000001 6, 35, 143, 391, 899, 1739, 3233, 5293, 8051, 11413, 17653,
> 24883, 33389, 43931, 56977, 72731, 92881, 118829, 145699, 176039
> %N A000001 Smallest prime between n^2 and (n+1)^2 times largest prime
> between n^2 and (n+1)^2.
> %C A000001 First 5 values are a subset of A132435.

This is always the case.
Let p and q be the smallest and largest primes in the interval [n^2,(n+1)^2].
Then

n^2 < p < q < (n+1)^2
n < sqrt(p) < sqrt(q) < n+1
sqrt(q) - sqrt(p) < n+1 - n = 1
0 < (sqrt(q) - sqrt(p))^2 = q - 2*sqrt(q*p) + p < 1
0 < q - 2*[sqrt(q*p)] + p < 3

Since q - 2*[sqrt(q*p)] + p is integer,
1 <= q - 2*[sqrt(q*p)] + p <= 2
[sqrt(q*p)] - p + 1 <= q - [sqrt(q*p)] <= [sqrt(q*p)] - p + 2.    (*)

Now let y = [sqrt(q*p)] - p + 1. Then
[sqrt(q*p)] - y = p - 1 and thus
nextprime([sqrt(q*p)] - y) = nextprime(p - 1) = p.
Furthermore, (*) implies that
q-1 <= [sqrt(q*p)] + y <= q
and thus nextprime([sqrt(q*p)] + y) = q.

Therefore, q*p belongs to A132435.

Regards,
Max