[seqfan] Re: New(?) sequence

[Allan Wechsler]

Still haring off on the idea of finding primes of each residue class of the
prime p.

Pick a prime, say, 13. Starting with 2, file the primes into their
appropriate residue classes modulo 13.  The best we could hope for is that
it would only take 13 primes to do this, but that doesn't happen. Instead,
we don't find a prime equal to 12 modulo 13 until we reach 103, the 27th
prime, and 12 is the last residue class of 13 to find a prime
representative. So it took 14 tries more than the minimum to find prime
representative of each of the residue classes of 13.

For 2, it takes 0 extra tries. For 3, it takes 1 extra try. For 5, it takes
3 extra tries (that is, the 8th prime fills out the last vacant residue
class).

Let A(n) = the number of consecutive primes we need to consider until
finding a representative of all the residue classes of p_n, minus p_n
(which is the minimum possible). The initial values of A(n) seem to be 0,
1, 3, 3, 3, 14, ... and this is not in OEIS.

Another idea would be closer to Zak Seidov's process, and look for
representatives starting not at 2 but at p_n itself. This is where Zak
noted that the number of extra tries for 7 is 0. The initial values of this
sequence are 0, 0, 1, 0, 6, 9, 4, ... and this isn't in OEIS either.

For each search regime we could also ask for the last residue class to be
filled in. If we always start searching at 2, the last classes are 1, 1, 4,
6, 10, 12, 1, ..., also not archived. Note that these are, so far,
always +1 or -1, but I doubt that pattern will persist.

If we start searching at p_n, the recalcitrant classes are  1, 1, 4, 1, 5,
12, 1, 1, ..., not archived.

This data is all from manual scratching, so I would really appreciate it if
somebody would confirm (and would be THRILLED if somebody could write some
code -- these are probably one-liners in Mathematica, Maple, or PARI).


On Mon, Jan 23, 2023 at 10:57 AM Allan Wechsler <acwacw at gmail.com> wrote:

> It's still kind of intriguing that:
>
> the two consecutive primes starting with 2 cover all residues modulo 2;
>
> the three consecutive primes starting with 3 cover all residues modulo 3;
>
> (5 fails);
>
> the seven consecutive primes starting with 7 cover all residues modulo 7;
>
> (11 fails) ...
>
> Does any other prime number meet this criterion? It seems unlikely.
>
>
> On Mon, Jan 23, 2023 at 2:24 AM <israel at math.ubc.ca> wrote:
>
>> On Jan 22 2023, zak seidov via SeqFan wrote:
>>
>> > {7,11,13,17,19,23} + n are primes for n =
>> > {0,90,16050,19410,43770,1091250,1615830,1954350,2822700,2839920}
>> and it continues 3243330, 3400200, 6005880, 6503580, 7187760, 7641360,
>> 8061990, 8741130, 10526550, 11086830, 11664540, 14520540, 14812860,
>> 14834700, 14856750, 16025820, 16094700, 18916470, 19197240, 19634040,
>> 19800360, 20112210, 20247030, 21321180, 21850170, 22587270, 24786390,
>> 25009410, 25524120, 27305550, 29153550, 31563930, 31875570, 32836740,
>> 33575940, 36319170, 36985290, 37055640, 40660710, 41214060, 41763420,
>> 41927430, 44842860, 45974550, 47204730, 48660240, 49157730, 50685900,
>> 50943780, 51255630, 53204850, 53266380, 55057890, 56431920, 57812460,
>> 59877390, 61052340, 62757960, 63655710, 65689560, 67022220, 69296310,
>> 72610110, 74283600, 74390070, 75085590, 76150500, 79413060, 82984530,
>> 87423090, 89483820, 94752720, ...
>>
>> More generally, you can have a similar sequence for primes
>> {p(j), p(j+1), ..., p(k)} as long as these don't cover all residues mod
>> any of p(j) to p(k).
>>
>> >29 + n's are not primes :(
>>
>> In case you missed it, that's because {7,11,13,17,19,23,29} cover all
>> residues mod 7, so at least one of {7,11,13,17,19,23,29} + n will be
>> divisible by 7.
>>
>> Cheers,
>> Robert
>>
>> >Zak Seidov
>> >zakseidov at yahoo.com
>> >
>> >--
>> >Seqfan Mailing list - http://list.seqfan.eu/
>> >
>> >
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>