Divisibility and Binomial Coefficients

[drew at math.mit.edu]

The below is a 3-digit analogue of the 2-digit njas sequence A053392
Add consecutive pairs of digits and concatenate them.  I used an
offset that distinguishes this from digital sum and A053392.

Add consecutive triplets of digits and concatenate them.
Offset 1000,2
Example: Where C means concatenate, a(1010) = (1+0+1)C(0+1+0) = 2C1 = 21.
a(1019) = (1+0+1)C(0+1+9) = 2C10 = 210. a(1020) = (1+0+2)C(0+2+0) =
32. a(1028) = (1+0+2)C(0+2+8) = 3C10 = 310. a(1029) = 3C11 = 311.
a(1030) = (1+0+3)C(0+3+0) = 4C3 = 43. a(1037) = (1+0+3)C(0+3+7) = 4C10
= 410.
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27,
28, 29, 210, 32, 33, 34, 35, 36, 37, 38, 39, 310, 311, 43, 44, 45, 46,
47, 48, 49, 410, 411, 412, 54, 55, 56, 57, 58, 59, 510, 511, 512, 513,
65, 66, 67, 68, 69, 610, ...

I notice that there are big chunks of the njas seq embedded in the new
seq.  For example the initial subsequence in the new seq
10,11,12,13,14,15,16,17,18,19,21,22,23,24,25,26,27,28,29,210 is also
A053392(100..114).

Again, the interesting question is the analogue of A053393 Periodic
points under the map (see A053392) that adds consecutive pairs of
digits and concatenates them. In this case, again, we have consecutive
triplets instead of consecutive pairs.

Does anyone want to code this and/or search for periodic points?

I think that the intricity of the njas seq A053393 is more than it
seems at first, when we see that as merely the first sequence in a
supersequence "Add consecutive n-tuples of digits and concatenate
them" and consider periodic points.

With a minute to go in the 3rd quarter of the decisive 7th game of the
Boston Celtics versus the Atlantic hawks, Boston dominates 79-41,
which I can't help but see as two primes.

happy Cinquo de Mayo tomorrow.

Best,

Jonathan Vos Post