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 (its 0)
9 says that the 9th digit of S is not 9 -- which is
true (its 1)
20 says that the
20th digit of S
is not 2 -- which is true (its 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 well 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 Neils OEIS.
_______________________
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