Multiplying over the comma

Hello Seqfans,

Here is how we’ll build S:

S = 1, ...

(b) to compute a(n+1), start by calling “d” the last digit of a(n-1); we have here:

d = 1

(c) call “i” the smallest integer which hasn’t yet multiplied “d”; we have here:

i = 1

(d) call P the product < d.i >; we have here:

P = d.i = 1.1 = 1

(e) let a(n+1) be the concatenation < i_P >; we have here:

a(n+1) = < i_P > = < 1_1 > = 11

Thus S goes now:

S = 1, 11, ...

(f) goto (b):

We have for the next term a(3):

d = 1 (the last digit of 11);

i = 2 [because “i” must be the smallest integer which hasn’t yet multiplied the digit “1” (the integer 1 was used before)];

P = d.i = 1.2 = 2

a(3) = i_P = 22

S = 1, 11, 22, ...

Again, goto (b):

We have for the next term a(4):

d = 2 (the last digit of the last term, 22);

i = 1 [because “i” must be the smallest integer which hasn’t yet multiplied the digit “2”];

P = d.i = 2.1 = 2

a(4) = i_P = 12

S = 1, 11, 22, 12, ...

Again, goto (b):

We have for the next term a(5):

d = 2 (the last digit of the last term, 12);

i = 2 [because “i” must be the smallest integer which hasn’t yet multiplied the digit “2”];

P = d.i = 2.2 = 4

a(5) = i_P = 24

S = 1, 11, 22, 12, 24, ...

Again, goto (b):

We have for the next term a(6):

d = 4 (the last digit of the last term, 24);

i = 1 [because “i” must be the smallest integer which hasn’t yet multiplied the digit “4”];

P = d.i = 4.1 = 4

a(6) = i_P = 14

S = 1, 11, 22, 12, 24, 14, ...

Again, goto (b):

We have for the next term a(7):

d = 4 (the last digit of the last term, 14);

i = 2 [because “i” must be the smallest integer which hasn’t yet multiplied the digit “4”];

P = d.i = 4.2 = 8

a(7) = i_P = 28

S = 1, 11, 22, 12, 24, 14, 28, ...

Again, goto (b):

We have for the next term a(8):

d = 8 (the last digit of the last term, 28);

i = 1 [because “i” must be the smallest integer which hasn’t yet multiplied the digit “8”];

P = d.i = 8.1 = 8

a(8) = i_P = 18

S = 1, 11, 22, 12, 24, 14, 28, 18, ...

Again, goto (b):

We have for the next term a(9):

d = 8 (the last digit of the last term, 18);

i = 2 [because “i” must be the smallest integer which hasn’t yet multiplied the digit “8”];

P = d.i = 8.2 = 16

a(9) = i_P = 216

S = 1, 11, 22, 12, 24, 14, 28, 18, 216, ...

Again, goto (b):

We have for the next term a(10):

d = 6 (the last digit of the last term, 216);

i = 1 [because “i” must be the smallest integer which hasn’t yet multiplied the digit “6”];

P = d.i = 6.1 = 6

a(10) = i_P = 16

S = 1, 11, 22, 12, 24, 14, 28, 18, 216, 16, ...

We see that S goes smoothly like this:

S = 1, 11, 22, 12, 24, 14, 28, 18, 216, 16, 212, 48, 324, 312, 510.

... but here we have to introduce a new rule – else will stop (because of zero, the last digit of 510):

(e2) “i” has to be discarded if the product d.i ends with zero; try the next smallest “i” (and never come back to the “bad” “i”).

We see at the end here... that the 0 comes from the 5;

S = 1, 11, 22, 12, 24, 14, 28, 18, 216, 16, 212, 48, 324, 312, 510.

... if we replace 5 by 6 (the next “i”), everything will be fixed:

S = 1, 11, 22, 12, 24, 14, 28, 18, 216, 16, 212, 48, 324, 312, 612, ...

All terms of S will be different, by construction – and S is quite fun to watch: multiplying two terms is like considering that the comma < , > is the multiplying sign < x >. The full operation (multiplication and result) is under the very eyes of the reader – and only that, no parasite characters!

If this is of interest, this could be extended quite easily and submitted to the OEIS:

S = 1,11,22,12,24,14,28,18,216,16,212,36,318,324,312,48,432,612,714,416,424,624,728,...

1,11,22,12,24,14,28,18,216,16,212,48,324,312,612,714,416,318,432, ...

Best,

É.

__________

Breaking news:

Maximilian Hasler has corrected and extended the sequence S here.

Many thanks!

Best,

É.