# [seqfan] Re: 2D Flavius's sieve

Neil Sloane njasloane at gmail.com
Thu Nov 26 08:40:06 CET 2020

```those sequences sound interesting!

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Email: njasloane at gmail.com

On Wed, Nov 25, 2020 at 6:43 PM Elijah Beregovsky <
elijah.beregovsky at gmail.com> wrote:

> I've come up with three new sequences akin to the Flavius's sieve A000960
> <https://oeis.org/A000960>, but in 2D.
> First, we number an infinite board diagonally (as in A316588
> <https://oeis.org/A316588>). Then we delete every second diagonal. We're
> left with {1,4,5,6,11,12,13,14,15,22,23...}. Then we renumber the board
> with the numbers that survived the previous step and delete every third
> diagonal. We're left with {1,4,5,13,14,15,22,23,24,25,26,27,42,43...}. Then
> renumber again and delete every fourth diagonal, and so on. What's left is
> our sequence. Another version deletes rows and another one deletes columns.
> Here's the mathematica code:
>
> dynP[l_, p_] :=
>   MapThread[l[[# ;; #2]] &, {{0}~Join~Most@# + 1, #} &@Accumulate at p];
> Trngl := dynP[#, Range[(-1 + Sqrt[1 + 8*Length[#]])/2]] &;
> lst = Trngl[Range[100]]; i = 2;
> While[Length[lst] > i,
>   lst = Trngl[Flatten[Drop[lst, {i, -1, i}]]];
>   i++];
> lst = Flatten[lst]
>
> It returns {1, 4, 5, 13, 14, 15, 26, 27, 42, 43, 44, 45, 56, 57, 58}.
>
> To get the other two sequences we change lst = Trngl[Flatten[Drop[lst, {i,
> -1, i}]]]; to
>
> lst = Trngl[Flatten[DeleteCases[Drop[PadRight[lst], None, {i, -1, i}], 0,
> Infinity]]]
>
> to yield {1, 2, 4, 6, 7, 11, 13, 16, 18, 24, 26, 28, 33, 35, 41} and to
>
> lst = Trngl[Flatten[DeleteCases[Drop[PadLeft[lst], None, {-i, 1, -i}], 0,
> Infinity]]]
>
> to yield {1, 3, 4, 8, 10, 11, 17, 19, 21, 24, 32, 36, 37, 39, 45}.
>
> None of these sequences are in OEIS. Are they interesting? Could you find
> an asymptotic formula for them? The log-log plots for the first ten million
> terms look convincingly like a line with a slope of 3/2, so I conjecture
> that a(n)~n^(3/2), but I can't prove it, nor have I run a regression on
> these values. Would be grateful for any ideas and contributions.