First digit of a(n)

... is not the a(n)th digit of S

(this message appeared more or less like this

on the Seqfan Mailing list on February 21st, 2010)

Hello Seqfans,

... computed by hand, here is the first (I guess) lexicographically reordering S of the Natural numbers where every a(n) says:

- "My first digit is _not_ the a(n)th digit of S":

S = 2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 20, 22, 23, 24, 11, 13, 14, 15, 16, 12, 17, 18, 19, 21, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, ...

Example:

10 says that the 10th digit of S is not 1 -- which is true (its 0)

9 says that the  9th digit of S is not 9 -- which is true (its 1)

20 says that the 20th digit of S is not 2 -- which is true (its 1)

...

The building method is simple: we must prolong S with the smallest unused integer which has no digit matching the corresponding digit of _the 3rd line_ of the array below. In the 4th line, read vertically means that the digit count must be red vertically -- in grey or bold. For example, this portion of the array means:

|12| <-- herunder is the 12th integer of S

,22, <-- this is the 12th integer of S

11  <-- 12th and

23  <-- 13th digit of S - here the 12th digit of S is 2 and the 13th is also 2)

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|31|32|33|34|35|36|37|...

-----> S = 2,1,4,3,6,5,8,7,10,9, 20,22,23,24,11,13,14,15,16,12,17,18,19,21,25,26,27,28,29,30,31,32,33,34,35,37,38,...

S digit# = 1 2 3 4 5 6 7 8 91 1  11 11 11 11 22 22 22 22 22 33 33 33 33 33 44 44 44 44 44 55 55 55 55 55 66 66 66 ...

S digit# = read.vertically .0 1  23 45 67 89 01 23 45 67 89 01 23 45 67 89 01 23 45 67 89 01 23 45 67 89 01 23 45 ...

_____

Ray Chandler has just checked the S sequence above and computed 100 terms; here they are:

S = 2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 20, 22, 23, 24, 11, 13, 14, 15, 16, 12, 17, 18, 19, 21, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 36, 41, 42, 43, 44, 45, 46, 47, 49, 50, 48, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 77, 78, 79, 80, 82, 83, 84, 85, 86, 87, 88, 89, 90, 92, 93, 94, 95, 96, 97, 98, 99, 200, 202, 203, 204, 205,...

__________

BTW:

Two different sequences based on the same idea (but dropping the re-ordering of the Naturals constraint from above):

T is the first lexicographically and monotonically increasing seq. where the 1st digit of a(n) is not the a(n)th digit of T:

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|31|32|33|34|35|36|...

T = a(n) = 2,3,4,5,6,7,8,9,10,12,14,16,18,20,23,30,31,32,34,40,41,42,43,45,50,51,52,53,54,60,61,62,63,64,70,71,...

T digit# = read.vertically .1 11 11 11 11 12 22 22 22 22 23 33 33 33 33 34 44 44 44 44 45 55 55 55 55 56 66 66 ...

T digit# = 1 2 3 4 5 6 7 8 90 12 34 56 78 90 12 34 56 78 90 12 34 56 78 90 12 34 56 78 90 12 34 56 78 90 12 34 ...

Example:

for n=1, a(n)=2 whose 1st digit 2 is not the #2 digit of T -- which is a 3

for n=2, a(n)=3 whose 1st digit 3 is not the #3 digit of T -- which is a 4

for n=3, a(n)=4 whose 1st digit 4 is not the #4 digit of T -- which is a 5

...

for n=8, a(n)=9 whose 1st digit 9 is not the #9 digit of T -- which is a 1 -- the 1 from 10

for n=9, a(n)=10 whose 1st digit 1 is not the #10 digit of T -- which is a 0 -- the 0 from 10

for n=10, a(n)=12 whose 1st digit 1 is not the #12 digit of T -- which is a 2 -- the 2 from 12

for n=11, a(n)=14 whose 1st digit 1 is not the #14 digit of T -- which is a 4 -- the 4 from 14

...

for n=20, a(n)=40 whose 1st digit 4 is not the #40 digit of T -- which is a 5 -- the 5 from 45

... so the 1st digit of a(n) is never the a(n)th digit of T.

We see here that 11 is forbidden in T; 11 would momentarily replace 12 in T -- but then 11, saying the 11th digit of T is not a 1 would lie -- as the 11th digit of T would be 1, precisely -- the 1 of 11. Putting 12 instead of 11 moves the cursor towards the 2 of 12 -- which digit is indeed different from the 1 of 12. As T is monotonically increasing, 11 will never appear because of the presence of 12.

_____

Ray Chandler has just corrected my T sequence above and computed 100 terms; here they are:

T = 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 23, 30, 31, 32, 34, 40, 41, 42, 43, 45, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 70, 71, 72, 73, 74, 80, 81, 82, 83, 84, 90, 91, 92, 93, 94, 100, 101, 102, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 300, 301, 303, 304, 305, 306, 307, 308, 309, 310, 311, 312, 313, 314,...

_____

If we drop the monotonically increasing requirement of T, we get U:

U is the first lexicographically seq. where the 1st digit of a(n) is not the a(n)th digit of U:

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|31|32|33|34|35|36|...

U = a(n) = 2,1,4,3,6,5,8,7,10, 9,11,15,17,19,21,23,25,27,29,31,34,40,33,35,39,41,43,45,47,49,50,51,53,55,57,60,...

U digit# = read.vertically .1  1 11 11 11 11 22 22 22 22 22 33 33 33 33 33 44 44 44 44 44 55 55 55 55 55 66 66 ...

U digit# = 1 2 3 4 5 6 7 8 90  1 23 45 67 89 01 23 45 67 89 01 23 45 67 89 01 23 45 67 89 01 23 45 67 89 01 23 ...

Example:

for n=1, a(n)=2 whose 1st digit 2 is not the #2 digit of U -- which is a 1

for n=2, a(n)=1 whose 1st digit 1 is not the #1 digit of U -- which is a 2

for n=3, a(n)=4 whose 1st digit 4 is not the #4 digit of U -- which is a 3

...

for n=8, a(n)=7 whose 1st digit 7 is not the #7 digit of U -- which is a 8

for n=9, a(n)=10 whose 1st digit 1 is not the #10 digit of U -- which is a 0 -- the 0 from 10

for n=10, a(n)=9 whose 1st digit 9 is not the #9 digit of U -- which is a 7

for n=11, a(n)=11 whose 1st digit 1 is not the #11 digit of U -- which is a 9

...

for n=20, a(n)=31 whose 1st digit 3 is not the #31 digit of U -- which is a 1 -- the 1 from 31

... so the 1st digit of a(n) is never the a(n)th digit of U.

We see here that 12 and 13, for instance, will never appear in U -- as 12 or 13 say that the 12th or 13th digit of U _are not_ 1, which is false: the 12th and the 13th digit of U are both 1 -- coming from 11.

_____

Ray Chandler has just checked the U sequence above and computed 100 terms; here they are:

U = 2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 11, 15, 17, 19, 21, 23, 25, 27, 29, 31, 34, 40, 33, 35, 39, 41, 43, 45, 47, 49, 50, 51, 53, 55, 57, 60, 61, 63, 65, 67, 69, 70, 72, 73, 75, 77, 79, 80, 81, 82, 83, 84, 85, 87, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 123, 124, 126, 129, 130, 132, 133, 135, 136, 138, 139, 141, 142, 144}

_____

And well get three more sequences like this if we replace first by last, in the definition; this gives:

Every a(n) says: "My last digit is _not_ the a(n)th digit of V"

Here is a re-ordering V of the Naturals obeying the rule « The last digit of a(n) is _not_ the a(n)th digit of V »:

-----> V = 2,1,4,3,6,5,8,7,11,9,10,12,13,14,15,16,17,18,20,19,30,21,22,23,24,31,26,25,27,28,32,29,33,34,35,36,...

V digit# = read.vertically .1 1 11 11 11 11 22 22 22 22 22 33 33 33 33 33 44 44 44 44 44 55 55 55 55 55 66 66 ...

V digit# = 1 2 3 4 5 6 7 8 90 1 23 45 67 89 01 23 45 67 89 01 23 45 67 89 01 23 45 67 89 01 23 45 67 89 01 23 ...

The building method is simple: we must prolong V with the smallest unused integer which has no digit matching the corresponding digit of _the last line_ of the above array (the line below the line with read vertically).

_____

Ray Chandler has checked the V sequence above and computed 100 terms; here they are:

V = 2, 1, 4, 3, 6, 5, 8, 7, 11, 9, 10, 12, 13, 14, 15, 16, 17, 18, 20, 19, 30, 21, 22, 23, 24, 31, 26, 25, 27, 28, 32, 29, 33, 34, 35, 36, 37, 38, 40, 39, 41, 50, 42, 43, 44, 45, 51, 46, 47, 48, 49, 52, 53, 54, 55, 56, 57, 58, 60, 59, 61, 62, 70, 63, 64, 65, 66, 71, 67, 68, 69, 72, 73, 74, 75, 76, 77, 78, 90, 79, 80, 81, 82, 91, 83, 84, 86, 85, 92, 87, 88, 89, 93, 94, 95, 96, 97, 98, 101, 99,...

Any taker for the last two sequences, W and X?

W: we drop the re-ordering of the Naturals but we want W to be monotonically increasing

X: we drop both the re-ordering of the Naturals and the monotonically increasing rule.

Ray has computed both sequences (100 terms); here they are:

W = 2, 3, 4, 5, 6, 7, 8, 9, 11, 20, 21, 22, 23, 24, 25, 26, 27, 29, 31, 33, 40, 41, 42, 43, 44, 45, 47, 49, 51, 53, 55, 60, 61, 62, 63, 65, 67, 69, 71, 73, 75, 77, 80, 81, 83, 85, 87, 89, 100, 101, 102, 103, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 28, 229, 230, 231, 232, 233, 234, 235, 236, 300, 301, 302, 304, 305, 306, 307, 400, 401, 402, 403,...

X = 2, 1, 4, 3, 6, 5, 8, 7, 11, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 30, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 48, 49, 50, 52, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 100, 101, 102, 103, 104, 105, 107, 108, 109,...

__________

P.-S.

Working on this page suddenly makes me think of a nice (?) Y sequence:

« First lexically re-ordering of the Naturals where neither a(n)th first or a(n)th last digit are the a(n)th digit of Y »

We just have to combine the building methods of S and V to get Y  no digit of Y can match any bold or grey digit below it:

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|31|32|33|34|35|36|37|...

-----> Y = 2,1,4,3,6,5,8,7,12, 9,30,20,22,23,10,11,13,14,15,16,17,18,19,21,25,31,26,28,27,24,32,29,33,34,35,37,38,...

Y digit# = 1 2 3 4 5 6 7 8 91  1 11 11 11 11 22 22 22 22 22 33 33 33 33 33 44 44 44 44 44 55 55 55 55 55 66 66 66 ...

Y digit# = read.vertically .0  1 23 45 67 89 01 23 45 67 89 01 23 45 67 89 01 23 45 67 89 01 23 45 67 89 01 23 45 ...

_____

Ray Chandler has corrected my Y sequence above and computed 100 terms; here they are:

Y = 2, 1, 4, 3, 6, 5, 8, 7, 12, 9, 30, 20, 22, 23, 10, 11, 13, 14, 15, 16, 17, 18, 19, 21, 25, 31, 26, 28, 27, 24, 32, 29, 33, 34, 35, 37, 38, 39, 40, 36, 41, 50, 42, 43, 44, 45, 51, 46, 47, 48, 52, 53, 54, 55, 49, 56, 57, 58, 60, 59, 62, 63, 70, 64, 65, 66, 67, 72, 68, 69, 74, 73, 75, 76, 77, 78, 79, 80, 90, 82, 84, 83, 85, 92, 86, 87, 88, 89, 93, 94, 95, 96, 98, 97, 99, 300, 200, 202, 204, 203,...

Ray adds two more sequences, Y1 and Y2;

_____

This suggests the ultimate Z sequence (many thanks to Franklin T. Adams-Watters and William Keith for their interesting remarks  see the Seqfan archives around February 21st, 2010):

« No digit of a(n) is the a(n)th digit of Z »

Ray Chandler computed 100 terms  which diverge from Y at Z(59)=62 (instead of 60  see bold figure 60 above):

Z = 2, 1, 4, 3, 6, 5, 8, 7, 12, 9, 30, 20, 22, 23, 10, 11, 13, 14, 15, 16, 17, 18, 19, 21, 25, 31, 26, 28, 27, 24, 32, 29, 33, 34, 35, 37, 38, 39, 40, 36, 41, 50, 42, 43, 44, 45, 51, 46, 47, 48, 52, 53, 54, 55, 49, 56, 57, 58, 62, 59, 60, 63, 70, 64, 65, 66, 67, 73, 68, 69, 72, 74, 75, 76, 77, 78, 79, 80, 90, 82, 84, 83, 86, 92, 85, 87, 88, 89, 93, 94, 95, 96, 98, 202, 97, 99, 200, 203, 204, 205,...

The Z sequence is no re-ordering of the Naturals as the integer 1023456789 will never appear...

___________

Now Ray adds four more sequences  Y2, Y3, Z2, Z3:

>Y2, Y3 are the variations of Y as T and U are to S.

>Z2, Z3 are the variations of Z as T and U are to S.

Y2 = 2, 3, 4, 5, 6, 7, 8, 9, 11, 20, 21, 22, 23, 24, 30, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 60, 61, 62, 63, 64, 65, 66, 67, 70, 71, 72, 80, 81, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 300, 301, 303, 304, 305, 306, 307, 400, 401, 410, 411, 412, 413, 414,...

Y3 = 2, 1, 4, 3, 6, 5, 8, 7, 11, 9, 14, 20, 13, 15, 17, 19, 21, 22, 23, 25, 29, 30, 31, 33, 35, 37, 40, 41, 42, 43, 45, 47, 49, 50, 51, 52, 55, 56, 57, 59, 60, 62, 63, 64, 66, 68, 69, 70, 71, 72, 73, 74, 75, 77, 78, 79, 80, 81, 82, 84, 85, 86, 87, 88, 89, 90, 91, 92, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 109, 110, 111, 112, 113, 114, 115, 116, 119, 120, 121, 122, 123, 124, 126, 127, 128, 129, 130,...

Z2 = 2, 3, 4, 5, 6, 7, 8, 9, 11, 20, 21, 22, 23, 24, 30, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 60, 61, 62, 63, 64, 65, 66, 67, 70, 71, 72, 80, 81, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 311, 313, 314, 330, 333, 334, 335, 400, 401, 410, 411, 412, 413, 414,...

Z3 = 2, 1, 4, 3, 6, 5, 8, 7, 11, 9, 14, 20, 13, 15, 17, 19, 21, 22, 23, 25, 29, 30, 31, 33, 35, 37, 40, 41, 42, 43, 45, 47, 49, 50, 51, 52, 55, 56, 57, 59, 60, 62, 63, 64, 66, 68, 69, 70, 71, 72, 73, 74, 75, 77, 78, 79, 80, 81, 82, 84, 85, 86, 87, 88, 89, 90, 91, 92, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 106, 109, 110, 111, 112, 113, 114, 115, 116, 119, 120, 121, 122, 123, 124, 126, 128, 129, 130, 131, 132,...

Many thanks to all the contributors! -- Those seq. will be submitted soon to Neils OEIS.

_______________________

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