Self-building the decimal expansion of Pi

with a(a(n))

 

Hello SeqFans,

 

this is A000796 (decimal expansion of Pi):

 

3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3, 3, 8, 3, 2, 7, 9, 5, 0, 2, 8, 8, 4, 1, 9, 7, 1, 6, 9, 3, 9, 9, 3, 7, 5, 1, 0, 5, 8, 2, 0, 9, 7, 4, 9, 4, 4, 5, 9, 2, 3, 0, 7, 8, 1, 6, 4, 0, 6, 2, 8, 6, 2, 0, 8, 9, 9, 8, 6, 2, 8, 0, 3, 4, 8, 2, 5, 3, 4, 2, 1, 1, 7, 0, 6, 7, ...

 

At every step, the sequence T hereunder produces a new chunk of Pi digits.

When concatenated and provided with the necessary commas, these chunks reproduce A000796.

The elegant (!) definition/formula for T is: « a(a(n)) is the nth chunk of Pi digits ».

 

T = 4, 3, 1, 3, 5, 26, 6, 5, 2, 8, 10, 26, 19, 26, 2, 31, 10, 1, 7, 6, 1, 2, 2, 10, 2, 9, 19, 26, 5, 0, 23, 10, 10, 1, 13, 19, 3, 7, 26, 2, 26, 26, 2, 19, 5, 11, 5,...

 

From now on, T is simply produced by A000796 (in short, A) thanks to this “translating table”:

 

0 in A is 0 in T

1 in A is 3 in T

2 in A is 9 in T

3 in A is 2 in T

4 in A is 1 in T

5 in A is 5 in T

6 in A is 7 in T

7 in A is 19 in T

8 in A is 10 in T

9 in A is 26 in T

 

Example; in the decimal expansion of Pi, the four yellow integers 5, 1, 0, 5 were encoded in T by the three yellow integers 5, 11, 5 (when you insert 5 in the formula, you get 5; when you insert 11, you get ‘10’ -- which needs a comma between 1 and 0):

 

A = 3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3, 3, 8, 3, 2, 7, 9, 5, 0, 2, 8, 8, 4, 1, 9, 7, 1, 6, 9, 3, 9, 9, 3, 7, 5, 1, 0, 5, 8, 2, 0, 9, 7, 4, 9, 4, 4, 5, 9, 2, 3, 0, 7, 8, 1, 6, 4, 0, 6, 2, 8, 6, 2, 0, 8, 9, 9, 8, 6, 2, 8, 0, 3, 4, 8, 2, 5, 3, 4, 2, 1, 1, 7, 0, 6, 7, ...

 

T = 4, 3, 1, 3, 5, 26, 6, 5, 2, 8, 10, 26, 19, 26, 2, 31, 10, 1, 7, 6, 1, 2, 2, 10, 2, 9, 19, 26, 5, 0, 23, 10, 10, 1, 13, 19, 3, 7, 26, 2, 26, 26, 2, 19, 5, 11, 5,...

 

From now on, using the translating table, we extend T like this:

 

A = (...)5, 1, 0, 5,  8, 2, 0,  9,  7, 4,  9, 4, 4, 5,  9, 2, 3, 0,  7,  8, 1, 6, 4, 0, 6, 2,  8, 6, 2, 0,  8,  9, ...

T = (...)  5, 11, 5, 10, 9, 0, 26, 19, 1, 26, 1, 1, 5, 26, 9, 2, 0, 19, 10, 3, 7, 1, 0, 7, 9, 10, 7, 9, 0, 10, 26, ...

 

One last thing must be said: T invisibly starts with a 0 (zero) -- and this 0 corresponds to the cases where a(n) = 0. This is forced by the digits 0 in Pi itself.

 

To see how the formula a(a(n)) works, here are the successive chunks of Pi for n = 1 to 47 (only five yellow chunks of Pi have more than one digit) :

 

Pi = 3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5,  8,  9,  7,  9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3, 3,  8, 3, 2,  7,  9, 5, 0

 n = 1  2  3  4  5  6    7   8  9 10  11  12  13  14 15   16  17 18 19   20  21 22 23  24 25 26  27  28 29 30

 T = 4, 3, 1, 3, 5, 26,  6,  5, 2, 8, 10, 26, 19, 26, 2,  31, 10, 1, 7,  6,   1, 2, 2, 10, 2, 9, 19, 26, 5, 0,

 

Pi =  2,  8,  8, 4, 1, 9, 7, 1, 6,  9, 3,  9,  9, 3,  7, 5, 1, 0, 5,  8, 2, 0,  9, ...

 n = 31  32  33 34   35  36 37 38  39 40  41  42 43  44 45   46  47  48 49 50  51  ...

 T = 23, 10, 10, 1,  13, 19, 3, 7, 26, 2, 26, 26, 2, 19, 5,  11,  5, 10, 9, 0, 26, ...

 

And this is how T must be red to reproduce Pi:

 

For n= 1 the  1st chunk of digits of Pi is to be seen under n= 4 [which is 3]

For n= 2 the  2nd chunk of digits of Pi is to be seen under n= 3 [which is 1]

For n= 3 the  3rd chunk of digits of Pi is to be seen under n= 1 [which is 4]

For n= 4 the  4th chunk of digits of Pi is to be seen under n= 3 [which is 1]

For n= 5 the  5th chunk of digits of Pi is to be seen under n= 5 [which is 5]

For n= 6 the  6th chunk of digits of Pi is to be seen under n=26 [which is 9]

For n= 7 the  7th chunk of digits of Pi is to be seen under n= 6 [which is 26]

For n= 8 the  8th chunk of digits of Pi is to be seen under n= 5 [which is 5]

For n= 9 the  9th chunk of digits of Pi is to be seen under n= 2 [which is 3]

For n=10 the 10th chunk of digits of Pi is to be seen under n= 8 [which is 5]

For n=11 the 11th chunk of digits of Pi is to be seen under n=10 [which is 8]

For n=12 the 12th chunk of digits of Pi is to be seen under n=26 [which is 9]

For n=13 the 13th chunk of digits of Pi is to be seen under n=19 [which is 7]

For n=14 the 14th chunk of digits of Pi is to be seen under n=26 [which is 9]

For n=15 the 15th chunk of digits of Pi is to be seen under n= 2 [which is 3]

For n=16 the 16th chunk of digits of Pi is to be seen under n=31 [which is 23]

For n=17 the 17th chunk of digits of Pi is to be seen under n=10 [which is 8]

For n=18 the 18th chunk of digits of Pi is to be seen under n= 1 [which is 4]

For n=19 the 19th chunk of digits of Pi is to be seen under n= 7 [which is 6]

For n=20 the 20th chunk of digits of Pi is to be seen under n= 6 [which is 26]

For n=21 the 21st chunk of digits of Pi is to be seen under n= 1 [which is 4]

For n=22 the 22nd chunk of digits of Pi is to be seen under n= 2 [which is 3]

For n=23 the 23rd chunk of digits of Pi is to be seen under n= 2 [which is 3]

For n=24 the 24th chunk of digits of Pi is to be seen under n=10 [which is 8]

For n=25 the 25th chunk of digits of Pi is to be seen under n= 2 [which is 3]

For n=26 the 26th chunk of digits of Pi is to be seen under n= 9 [which is 2]

For n=27 the 27th chunk of digits of Pi is to be seen under n=19 [which is 7]

For n=28 the 28th chunk of digits of Pi is to be seen under n=26 [which is 9]

For n=29 the 29th chunk of digits of Pi is to be seen under n= 5 [which is 5]

For n=30 the 30th chunk of digits of Pi is to be seen under n= 0 [which is 0]

For n=31 the 31st chunk of digits of Pi is to be seen under n=23 [which is 2]

For n=32 the 32nd chunk of digits of Pi is to be seen under n=10 [which is 8]

For n=33 the 33rd chunk of digits of Pi is to be seen under n=10 [which is 8]

For n=34 the 34th chunk of digits of Pi is to be seen under n= 1 [which is 4]

For n=35 the 35th chunk of digits of Pi is to be seen under n=13 [which is 19]

For n=36 the 36th chunk of digits of Pi is to be seen under n=19 [which is 7]

For n=37 the 37th chunk of digits of Pi is to be seen under n= 3 [which is 1]

For n=38 the 38th chunk of digits of Pi is to be seen under n= 7 [which is 6]

For n=39 the 39th chunk of digits of Pi is to be seen under n=26 [which is 9]

For n=40 the 40th chunk of digits of Pi is to be seen under n= 2 [which is 3]

For n=41 the 41st chunk of digits of Pi is to be seen under n=26 [which is 9]

For n=42 the 42nd chunk of digits of Pi is to be seen under n=26 [which is 9]

For n=43 the 43rd chunk of digits of Pi is to be seen under n= 2 [which is 3]

For n=44 the 44th chunk of digits of Pi is to be seen under n=19 [which is 7]

For n=45 the 45th chunk of digits of Pi is to be seen under n= 5 [which is 5]

For n=46 the 46th chunk of digits of Pi is to be seen under n=11 [which is 10]

For n=47 the 47th chunk of digits of Pi is to be seen under n= 5 [which is 5]

    ...                                     this column forms T, and this Pi (with the necessary commas)

 

 

My challenge in building T was to minimize the quantity of “yellow chunks” (only five of them), and the size of the “translating table” (the largest figure in the table is 26). Can this be improved?

 

Best,

É.