[seqfan] Re: A090466

[Chris Thompson]

Picking up a thread you probably thought dead...

Using a slightly different notation from that used in A090466,
the n'th (k+2)-gonal number is P(n,k) = [n(n-1)/2]k + n, and
A090466 is the union of the P(n,k) for n>=3, k>=1. I will refer
to P(n,k) as being in the n'th row and k'th column of 2-D array.

On Jan 19 2017, Charles Greathouse wrote:

>I agree, (number of terms up to x)/x should converge. The limit could be
>interpreted as the probability that a random number is a nontrivial
>figurate number. A first approximation of the limit is
>
>1 - product_{n >= 3} 1 - 2/(n^2 - n) = 2/3 = 0.6666...

That would give the right answer if the rows were behaving
independently, i.e. the triangular numbers n(n-1)/2 were pairwise
coprime. A pity that is not the case, especially given the
nice rational value for the product!

>and a better approximation is
>
>1 - product_{n >= 3, n ≠ 6} 1 - 2/(n^2 - n) = 9/14 = 0.6428...
>
>noting that all hexagonal numbers are also triangular.

True that every hexagonal number is triangular: P(n,4) = P(2n-1,1),
but that's irrelevant as the P(n,k) for any fixed k contribute
an infinitesimal proportion of the numbers in A090466. It's also
true that the 6th row is a a subset of the 3rd row: P(6,k) = P(3,5k+1),
which would in a way justify dropping the 14/15 factor from the
product.

But the correlations between the rows are much more complicated than
that! The n'th row is a subset of the 3rd row for any n divisible by 3.
Also the 16th row is a subset of the 4th row, the 10th row is a subset
of the union of the 4th and 5th rows, etc. Meanwhile its easy to see
that the n'th and (n+1)'th rows are always disjoint.

A quick computation of the density of the union of first few rows gives

Including               Density
  3k + 3:        1/3       = 0.33333333...
  6k + 4:        1/2       = 0.50000000...
 10k + 5:       17/30      = 0.56666666...
 15k + 6:  no change
 21k + 7:      123/210     = 0.58571428...
 28k + 8:      251/420     = 0.59761904...
 36k + 9:  no change
 45k + 10: no change
 55k + 11:      20/33      = 0.60606060...
 66k + 12: no change
 78k + 13:     839/1365    = 0.61465201...
 91k + 14:    1237/2002    = 0.61788211...
105k + 15: no change
120k + 16: no change
136k + 17:  211451/340340  = 0.62129341...
153k + 18: no change
171k + 19: 4029093/6466460 = 0.62307553...
190k + 20: 3029627/4849845 = 0.62468532...
210k + 21: no change
231k + 22:  607051/969969  = 0.62584577...

Presumably the values are converging to the 0.64036-ish value
suggested by the comments in https://oeis.org/A090466 . How fast?
The contribution of the P(n,k) with n>N can be no more than

  sum_{n>N} 2 / n(n-1) = 2/N

Does it in fact behave like constant/N as N tends to infinity?

-- 
Chris Thompson
Email: cet1 at cam.ac.uk