[seqfan] Re: Comma numbers (2)

[Eric Angelini]

Aaaaarghllll :

> 78 is in S because we add to 45 the integer 34 seen around the comma of [23,45]

... no, 78 should be 79 :-(

I don't know now about "convergence" any more... I guess it might still happen...

You've got the general idea anyway
Sorry for my slow brain and computations
Best,
É.


-----Message d'origine-----
De : seqfan-bounces at list.seqfan.eu [mailto:seqfan-bounces at list.seqfan.eu] De la part de Eric Angelini
Envoyé : jeudi 15 octobre 2009 15:20
À : Sequence Fanatics Discussion list
Objet : [seqfan] Comma numbers


Hello SeqFans,

Let a(1) = 0
    a(2) = 1
and a(n) = a(n-1) + [the two-digit integer split by the comma
                     which separates a(n-1) and a(n-2)]

S = 0,1,1,12,23,45,78,135,216,268,330,413,417,451,525,540,595,600,656,662,728,755,...

23 is in S because we add to 12 the integer 11 seen around the comma of [1,12]
78 is in S because we add to 45 the integer 34 seen around the comma of [23,45]
417 is in S because we add to 413 the integer 4 seen around the comma of [330,413]

Note(1):

Those two different starts converge towards the same seq:

Sa = 4,16,57,122,193,214,246,288,350, ...
Sb = 4, 17, 58, 133, 214,246,288,350, ...

What are the laws ruling convergence?

Note(2):

I am looking for "comma numbers", which are numbers like [abc] where the split [a,bc]
or [ab,c] would produce later in the sequence the said "comma number" again: ..., abc, ...

Example: 416 is not a "comma number" because we have no hit for the two different
         starts [4,16] or [41,6]:

S1 = 4,16,57,122,193,214,246,288,350,433, ... <-- no hit
S2 = 41,6,22,84,112,153,174,205,247,299,371,464, ... <-- no hit

I guess 10 is the first "comma number":

10 --> 1,0,10,11,12,23, etc. --> '10' is in the seq
(we see that '20', '30', '40', etc. are in the seq too)

What about 11?

11 --> 1,1,12,23, etc. <-- no hit: '11' is not a "comma number"

How would the "comma numbers" seq look like?

Best,
É.

(this was inspired by:

A121805 The "commas" sequence: a(1) = 1; for n > 1, let x be the least significant digit of a(n-1); then a(n) = a(n-1) + x*10 + y where y is the most significant digit of a(n) and is the smallest such y, if such a y exists. If no such y exists, stop.




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