7
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Inspired by Interview Question or Pathbreaking puzzle and A121808.

Start with $1$, and count the number of times $1$ occurs, and report this in the format 'number of ones:1', i.e. the next term is $11$.

Repeating this, we get:

$1, 11, 12, 1121, 1321, 122131, 132231, 122232, 112431, 13213141, 14213241, 13223142, 12233241, 12233241, 12233241$

which is the OEIS sequence mentioned above.

Note that $12233241$ reports itself.

My question is:

Does any other number report itself?

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5
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I believe there is an implied restriction from JonMark Perry's comment on Glorfindel's answer that the 'number of Xs' has X going from 1 up to the largest digit in the number (i.e. the 1st, 3rd, 5th digits etc. form the sequence 1, 2, 3 etc.)

In this case, I have another solution:

1322334151 which has 3 1s, 2 2s, 3 3s, 1 4 and 1 5.

(NB: this solution works without the restriction, I just wanted to point out that it observes it.)

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Here is an exhaustive list, with the assumption that OP expects all digits to be present (IOW the resulting numbers all match a pattern like 1a2b3c4d... with a, b, c,... > 0).

TL;DR

2 digit numbers (1a): no solution
4 digit numbers (1a2b): no solution
6 digit numbers (1a2b3c): no solution
8 digit numbers (1a2b3c4d): 12233241 and 13213341
10 digit numbers (1a2b3c4d5e): 1322334151
12 digit numbers (1a2b3c4d5e6f): no solution
14 digit numbers (1a2b3c4d5e6f7g): 14233242516171
16 digit numbers (1a2b3c4d5e6f7g8h): 1523324152617181
18 digit numbers (1a2b3c4d5e6f7g8h9i): 162332415162718191

Two digit numbers (1a)

a can only be 2 (the total number of digits) which leads to an impossibility, therefore there is no solution

Four digit numbers (1a2b)

Forewords: below, the notation (n eq m) yields 1 if n equals m and 0 otherwise

We can state:
a = 1 + (a eq 1) + (b eq 1)
b = 1 + (a eq 2) + (b eq 2)
sum = a + b = 4 (total number of digits)

a = 1 leads to a paradox
a = 2 leads to a paradox too: b = 2 + (b eq 2)
Therefore there is no solution

Six digit numbers (1a2b3c)

We can state:
a = 1 + (a eq 1) + (b eq 1) + (c eq 1)
b = 1 + (a eq 2) + (b eq 2) + (c eq 2)
c = 1 + (a eq 3) + (b eq 3) + (c eq 3)
sum = a + b + c = 6 (total number of digits)

a = 1 leads to a paradox
a = 3 implies b = c = 1, which is impossible (sum = 6)
If a = 2, then b >= 2, therefore c = 1, which implies b = 3. This can only be possible if c = 2 which leads us to a paradox, therefore there is no solution

Eight digit numbers (1a2b3c4d)

We can state:
a = 1 + (a eq 1) + (b eq 1) + (c eq 1) + (d eq 1)
b = 1 + (a eq 2) + (b eq 2) + (c eq 2) + (d eq 2)
c = 1 + (a eq 3) + (b eq 3) + (c eq 3) + (d eq 3)
d = 1 + (a eq 4) + (b eq 4) + (c eq 4) + (d eq 4)
sum = a + b + c + d = 8 (total number of digits)

a = 1 leads to a paradox
a = 4 implies both d >= 2 and d = 1
b = 4 implies a = c = d = 2 which is impossible (sum = 8)
c = 4 implies implies a = b = d = 3 which is impossible (sum = 8)

Therefore d is 1 and we now have:
a = 2 + (b eq 1) + (c eq 1)
b = 1 + (a eq 2) + (b eq 2) + (c eq 2)
c = 1 + (a eq 3) + (b eq 3) + (c eq 3)
d = 1
If a = 2 then b >= 2 and c >= 2. b can only be 3 since b = 2 leads to a paradox. Thus c = 2 (sum = 8).
Therefore a solution is 12233241 (this was given by the OP)
If a = 3 then c >= 2 and b = 1, which is only possible if c = 3 since the sum of all digits is 8.
Therefore the only other solution is 13213341

Ten digit numbers (1a2b3c4d5e)

We can state:
a = 1 + (a eq 1) + (b eq 1) + (c eq 1) + (d eq 1) + (e eq 1)
b = 1 + (a eq 2) + (b eq 2) + (c eq 2) + (d eq 2) + (e eq 2)
c = 1 + (a eq 3) + (b eq 3) + (c eq 3) + (d eq 3) + (e eq 3)
d = 1 + (a eq 4) + (b eq 4) + (c eq 4) + (d eq 4) + (e eq 4)
e = 1 + (a eq 5) + (b eq 5) + (c eq 5) + (d eq 5) + (e eq 5)
sum = a + b + c + d + e = 10

a = 1 leads to a paradox
a = 5 implies b = c = d = e = 1 which doesn't fit sum
e <= 2, else if e >= 3, at least two of a,b,c,d=5 and the others are >= 1, which yields a total number of digits >= 5+5+1+1+3 = 15 which is higher than sum
d <= 2 else, similarly to method for e, digits >= 13 > sum
c <= 3 else digits >= 14 > sum
b <= 3 else digits >= 11 > sum
e = 2 implies either a = 1 (no) or a = 5 (not with e = 2) therefore e = 1

We now have:
a = 2 + (b eq 1) + (c eq 1) + (d eq 1)
b = 1 + (a eq 2) + (b eq 2) + (c eq 2) + (d eq 2)
c = 1 + (a eq 3) + (b eq 3) + (c eq 3)
d = 1 + (a eq 4)
e = 1

a = 4 implies d = 2, b = c = 1 which doesn't fit sum, therefore d = 1, a = 3, b >= 2 and c >= 2
c = 2 implies b = 2 which doesn't fit sum, therefore c = 3 and b = 2
Therefore there is only one solution: 1322334151 (found by boboquack)

Twelve digit numbers (1a2b3c4d5e6f)

We can state:
a = 1 + (a eq 1) + (b eq 1) + (c eq 1) + (d eq 1) + (e eq 1) + (f eq 1)
b = 1 + (a eq 2) + (b eq 2) + (c eq 2) + (d eq 2) + (e eq 2) + (f eq 2)
c = 1 + (a eq 3) + (b eq 3) + (c eq 3) + (d eq 3) + (e eq 3) + (f eq 3)
d = 1 + (a eq 4) + (b eq 4) + (c eq 4) + (d eq 4) + (e eq 4) + (f eq 4)
e = 1 + (a eq 5) + (b eq 5) + (c eq 5) + (d eq 5) + (e eq 5) + (f eq 5)
f = 1 + (a eq 6) + (b eq 6) + (c eq 6) + (d eq 6) + (e eq 6) + (f eq 6)
sum = a + b + c + d + e + f = 12

a = 1 leads to a paradox
a = 6 implies b = c = d = e = f = 1 which doesn't fit sum
f <= 2 else digits >= 18 > sum
e <= 2 else digits >= 16 > sum
d <= 2 else digits >= 14 > sum
c <= 3 else digits >= 15 > sum
b <= 4 else digits >= 14 > sum
f = 2 implies either a = 1 (no) or a = 6 (not with f = 2) therefore f = 1

We now have:
a = 2 + (b eq 1) + (c eq 1) + (d eq 1) + (e eq 1)
b = 1 + (a eq 2) + (b eq 2) + (c eq 2) + (d eq 2) + (e eq 2)
c = 1 + (a eq 3) + (b eq 3) + (c eq 3)
d = 1 + (a eq 4) + (b eq 4)
e = 1 + (a eq 5)
f = 1

a = 5 implies e = 2 and b = c = d = 1 which doesn't fit sum, therefore e = 1 and a >= 3
b = 4 implies a = c = d = 2 but a >= 3, therefore b <= 3
a = 4 implies d = 2, b = 3 (b = 2 impossible) and c = 2 which doesn't fit sum, therefore d = 1 and a >= 4

We now have:
a = 4 + (b eq 1) + (c eq 1)
b = 1 + (b eq 2) + (c eq 2)
c = 1 + (b eq 3) + (c eq 3)
d = 1
e = 1
f = 1

b = 3 is impossible (needs b eq 2), therefore c = 1
So, either b = 1 and a = 6, or b = 2 and a = 5 but in both cases, the total number of digits doesn't match the sum.
Therefore there is no solution.

Fourteen digit numbers (1a2b3c4d5e6f7g)

We can state:
(snipped, similar to above)
sum = a + b + c + d + e + f + g = 14

a = 1 leads to a paradox
a = 7 implies b = c = d = e = f = g = 1 which doesn't fit sum
g <= 2 else digits >= 21 > sum
f <= 2 else digits >= 19 > sum
e <= 2 else digits >= 17 > sum
d <= 2 else digits >= 15 > sum
c <= 3 else digits >= 16 > sum
b <= 4 else digits >= 15 > sum
g = 2 implies either a = 1 (no) or a = 7 (not with g = 2) therefore g = 1

We now have:
a = 2 + (b eq 1) + (c eq 1) + (d eq 1) + (e eq 1) + (f eq 1)
b = 1 + (a eq 2) + (b eq 2) + (c eq 2) + (d eq 2) + (e eq 2) + (f eq 2)
c = 1 + (a eq 3) + (b eq 3) + (c eq 3)
d = 1 + (a eq 4) + (b eq 4)
e = 1 + (a eq 5)
f = 1 + (a eq 6)
g = 1

a = 6 implies f = 2 and b = c = d = e = 1 which doesn't fit sum, therefore f = 1 and a >= 3
a = 5 implies e = 2 and b = c = d = 1 which doesn't fit sum, therefore e = 1, a = 4, b >= 2, c >= 2 and d >= 2

We now have:
a = 4
b = 1 + (b eq 2) + (c eq 2) + (d eq 2)
c = 1 + (b eq 3) + (c eq 3)
d = 2 + (b eq 4)
e = 1
f = 1
g = 1

Since d <= 2, then d = 2, b = 3 (b = 2 impossible) and c = 2, which leads to the only solution: 14233242516171

Sixteen digit numbers (1a2b3c4d5e6f7g8h)

We can state:
(snipped, similar to above)
a + b + c + d + e + f + g + h = 16

a = 1 leads to a paradox
h <= 2 else the total number of is > sum
h <= 2 else digits > sum
g <= 2 else digits > sum
f <= 2 else digits > sum
e <= 2 else digits > sum
d <= 3 else digits > sum
c <= 3 else digits > sum
b <= 5 else digits > sum
h = 2 implies a = 8 which is impossible with a and h > 1 therefore h = 1 and a >= 2

We now have:
a = 2 + (b eq 1) + (c eq 1) + (d eq 1) + (e eq 1) + (f eq 1) + (g eq 1)
b = 1 + (a eq 2) + (b eq 2) + (c eq 2) + (d eq 2) + (e eq 2) + (f eq 2) + (g eq 2)
c = 1 + (a eq 3) + (b eq 3) + (c eq 3) + (d eq 3)
d = 1 + (a eq 4) + (b eq 4)
e = 1 + (a eq 5) + (b eq 5)
f = 1 + (a eq 6)
g = 1 + (a eq 7)
h = 1

a = 7 implies g = 2 and b = c = d = e = f = 1 which doesn't fit sum, therefore g = 1 and a >= 3
a = 6 implies f = 2, b >= 3, c = d = e = 1 which contradicts b >= 3, therefore f = 1 and a >= 4
b = 5 implies a = c = d = e = 2 but a >= 4, therefore b <= 4

We now have:
a = 4 + (b eq 1) + (c eq 1) + (d eq 1) + (e eq 1)
b = 1 + (b eq 2) + (c eq 2) + (d eq 2) + (e eq 2)
c = 1 + (b eq 3) + (c eq 3) + (d eq 3)
d = 1 + (a eq 4) + (b eq 4)
e = 1 + (a eq 5)
f = 1
g = 1
h = 1

b = 4 implies c = d = e = 2, a = 4 and thus d = 3, therefore b <= 3
a = 4 implies both e = 1 and e != 1, therefore a = 5
a = 5 implies e = 2, d = 1, b = 3, c = 2, which leads to the only valid solution 1523324152617181

Eighteen digit numbers (1a2b3c4d5e6f7g8h9i)

We can state:
(snipped, similar to above)
a + b + c + d + e + f + g + h + i = 18

a = 1 leads to a paradox
i <= 2 else the total number of is > sum
h <= 2 else digits > sum g <= 2 else digits > sum f <= 2 else digits > sum e <= 2 else digits > sum d <= 4 else digits > sum c <= 4 else digits > sum b <= 5 else digits > sum i = 2 implies a = 9 which is impossible with a and i > 1 therefore i = 1 and a >= 2

We now have:
a = 2 + (b eq 1) + (c eq 1) + (d eq 1) + (e eq 1) + (f eq 1) + (g eq 1) + (h eq 1)
b = 1 + (a eq 2) + (b eq 2) + (c eq 2) + (d eq 2) + (e eq 2) + (f eq 2) + (g eq 2) + (h eq 2)
c = 1 + (a eq 3) + (b eq 3) + (c eq 3) + (d eq 3)
d = 1 + (a eq 4) + (b eq 4) + (c eq 4) + (d eq 4)
e = 1 + (a eq 5) + (b eq 5)
f = 1 + (a eq 6)
g = 1 + (a eq 7)
h = 1 + (a eq 8)
i = 1

a = 8 implies h = 2, b >= 3 and a <= 7, therefore h = 1, a >= 3
a = 7 implies g = 2, b >= 3, c = d = e = f = 1 which contradicts b >= 3, therefore g = 1, a >= 4
a = 6 implies f = 2, b >= 3, c >= 2, d = e = 1, therefore b = 3 and c = 2. This leads us to one solution: 162332415162718191

If a <= 5, then f = 1, thus a = 5 and b, c, d, e > 1, therefore e = 2, b >= 3, b <= 4 and we have:
a = 5
b = 2 + (c eq 2) + (d eq 2)
c = 1 + (b eq 3) + (c eq 3) + (d eq 3)
d = 1 + (b eq 4) + (c eq 4) + (d eq 4)
e = 2
f = 1
g = 1
h = 1
i = 1

b = 4 implies c = d = 2 but c = 2 implies one of b, c, d equals 3, therefore b = 3 and c >= 2
c = 4 impossible (needs c eq 3)
d = 4 impossible (needs b eq 4), therefore d = 1 which contradicts d > 1 (a = 5)
Therefore a cannot be <= 5 and we have found the only solution above.

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  • $\begingroup$ @JonMarkPerry I rollbacked your last edit which I didn't fully grasp. I gave details on my reasoning. Tell me if I'm missing something. $\endgroup$ – xhienne Sep 5 '18 at 10:12
  • $\begingroup$ I can see that now, my method involves putting e=4 into the (d) eqn. - at least 2 must be 4, etc... $\endgroup$ – JonMark Perry Sep 5 '18 at 10:22
  • $\begingroup$ which I can now see is garbage! $\endgroup$ – JonMark Perry Sep 5 '18 at 10:32
  • $\begingroup$ Anyway, it's nice you could have a look on such a long post and validate my findings. I'll soon come up with the remaining of that analysis, be patient. $\endgroup$ – xhienne Sep 5 '18 at 10:39
  • $\begingroup$ I checked up to 10, then I fell asleep! :). you can use hex to get to 32digits, although this introduces a new twist. $\endgroup$ – JonMark Perry Sep 5 '18 at 11:22
4
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Yes, there is.

22 (contains the number 2, 2 times)

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  • $\begingroup$ arguably goes to 1022 n'est-ce pas? $\endgroup$ – JonMark Perry Sep 4 '18 at 7:03
  • 3
    $\begingroup$ No, just like 1 goes to 11 and not 1120304050... $\endgroup$ – Glorfindel Sep 4 '18 at 7:04

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