[seqfan] Sums, chunks and divisions -- twice

Eric Angelini Eric.Angelini at kntv.be
Fri Nov 28 22:56:01 CET 2014

Hello SeqFans,
please set the width of your "reading window" to 90 characters without automatic "carriage return"-- else the explanations might be quite difficult to follow...
(or copy/paste the whole damn mail into Word -- then push the margins --equivalent and better pgms exists ).

Here is C:

C=1,2,4,3,5,6,7,8,9,71,10,11,12,13,14,15,16,17,19,18,20,21,41,22,23,24,25,33,26,...

This will ring a bell now, but wait:

a) Insert a ":" immediately after the 1st integer, then (from there) another ":" after 2 integers, then a ":" after 4 integers, then after 3 integers, etc. (the marks are inserted according to the terms of the sequence C itself); we get:

C=1:2,4:3,5,6,7:8,9,71:10,11,12,13,14:15,16,17,19,18,20:21,41,22,23,24,25,33:26,...

We see now a succession of :chunks: -- where the sum of the integers in each chunk is divisible by the first integer of the said chunk (let’s check with S=sum; D=division; Q=quotient; Z being the size of the chunk, given by C itself):
Z=1  2     4      3          5                 6                7            ..8..
C=1:2,4:3,5,6,7:8,9,71:10,11,12,13,14:15,16,17,19,18,20:21,41,22,23,24,25,33:26,...
S=1 +6   +21    +88       +60            +105                +182
D=1 6/2  21/3   88/8      60/10           105/15              182/21
Q=1 =3    =7    =11        =6               =7                  =8

This is not new (see my former post). But wait... Let’s start with C again:

C=1,2,4,3,5,6,7,8,9,71,10,11,12,13,14,15,16,17,19,18,20,21,41,22,23,24,25,33,26,...

b) Insert now a "*" immediately after the 1st _digit_, then (from there) another "*" after 2 _digits_, then "*" after 4 _digits_, then after 3 digits, etc. -- the stars are inserted according to the terms –not the digits!– of the sequence C itself. We get:

C=1*2,4*3,5,6,7*8,9,7*1,10,11*12,13,14*15,16,17,1*9,18,20,21,4*1,22,23,24,25,33*26,...

Miracle! We see again a succession of *chunks* -- where the sum of the _digits_ in each chunk is divisible by the first _digit_ of the said chunk (let’s check with S=sum; D=division; Q=quotient; Z being the size of the digit-chunk, given by C itself):

Z=1  2     4      3      5       6         7           8            9            ..71..
C=1*2,4*3,5,6,7*8,9,7*1,10,11*12,13,14*15,16,17,1*9,18,20,21,4*1,22,23,24,25*33,26,...
S=1 +6    +21    +24     +4     +12       +22         +27         +175
D=1 6/2   21/3   24/8   4/1    12/1      22/1        27/9         175/1
Q=1 =3    =7     =3     =4     =12       =22         =3           =175

C was always extended with the smallest integer not yet present in C, respecting both (a) and (b) -- and not forcing a _digit-chunk_ to start with zero (else a division by zero would arise)

Is C a permutation of the integers >0? I guess yes.

Best,
É.