# [seqfan] Re: Primes on the right line on the plane "n - prime(n)"

M. F. Hasler oeis at hasler.fr
Tue Apr 25 16:28:23 CEST 2017

```On Fri, Apr 21, 2017 at 3:13 AM, zak seidov via SeqFan
<seqfan at list.seqfan.eu> wrote:
> 20 primes on the right line on the plane "n - prime(n)":  y=197+6*(x-45).
> {n,prime(n)}:
> {{45, 197}, {51, 233}, {52, 239}, {54, 251}, {55, 257}, {56, 263}, {57, 269}, {64, 311}, {71, 353}, {72, 359}, {76, 383}, {77, 389}, {79, 401}, {86, 443}, {87, 449}, {89, 461}, {92, 479}, {94, 491}, {96, 503}, {97, 509}}.
> Worth submitting(?)
> Are there cases with  more primes on the right line?

Yes, among the first 5000 primes there are 25 on a straight line
prime(k+i) = prime(k)+m*i  with slope m=9,
31 on a line with slope m=8, and 52 on a line with slope m=10.

I think this is interesting, but I would suggest to rather submit only
the starting point of these sequences,
and the obvious generalizations:
a(n ; m) = least prime p=prime(k) such that there are
at least/exactly n primes of the form p(k+i) = p + m i, i >= 0
(your subset corresponding to (n,m) = (20, 6)).
Instead of the primes p one could also list the index k,
which I think is less useful but maybe justified from the encyclopedic
point of view.

Further, one could then also list the irregular tables of all primes
(instead of just the first one) having the respective property
(i.e., being the starting point of such a sequence).
In any case I would not submit just this particular example as a
finite sequence with 20 terms, but rather,
if it is really desired to list all the primes, as a row of an
irregular table having more examples in the other rows.
So your row could be the first one of a table listing all of the
(20,6) sequences (row lengths would all be 20, not sure whether there
would be many)
or row 20 of the m=6 table (most probably several preceding rows would
start with the same numbers in the "at least n" variant, so one should
prefer the "exactly n" variant).

- Maximilian

(PARI)
maxNumSame(v,x,c,m=[0])=for(i=1,#v=vecsort(v),if(v[i]==x,c++>m[1]&&m=[c,x],x=v[i];c=1));m
chk(m,N=5000)=maxNumSame(primes(N)-m*[1..N])
for(m=6,13,printf("m=%d: %d. ",m,chk(m)))

```