Eight plus five.
Exactly twenty-four letters.
Roughly twelve.
At least ten.
If you're counting letter tokens, then eighty-one, but if you're counting letter
types, then eighteen.
Just one more than are in the question to this answer!
Spelling puzzles
This sentence has _____ A's, _____ B's, _____ C's, _____ D's, _____ E's, _____ F's, _____ G's, _____ H's, _____ I's, _____ J, _____ K, _____ L's, _____ M's, _____ N's, _____ O's, _____ P's, _____ Q, _____ R's, _____ S's, _____ T's, _____ U's, _____ V's, _____ W's, _____ X's, _____ Y's, and _____ Z.
----------------------------------------- | 8 | 2 | 5 | 9 | 3 | 2 | 1 | 8 | 7 | 2 | ----------------------------------------- | | | | | | | | | | | -----------------------------------------In the first row there is the 10-digit number 8259321872. In the second row, the first square contains the digit (< 10) which shows the number of 0's in the 2 rows, the second square contains the digit which indicates the number of 1's in the 2 rows....., the last one shows the number of 9's. Field this table.
I 8 I 2 I 5 I 9 I 3 I 2 I 1 I 8 I 7 I 2 I I 1 I 6 I 5 I 1 I 0 I 2 I 1 I 1 I 2 I 1 I
I 4 I 9 I 4 I 2 I 1 I 4 I 9 I 8 I 9 I 4 I I 1 I 5 I 8 I 1 I 8 I 4 I 7 I 4 I 4 I 3 I I 1 I 8 I 1 I 1 I 9 I 1 I 0 I 1 I 4 I 4 I
I 0 I 0 I 0 I 0 I 0 I 0 I 0 I 0 I 8 I 8 I I 1 I 1 I 1 I 1 I 1 I 1 I 1 I 1 I 1 I 8 I I 2 I 2 I 2 I 2 I 2 I 2 I 2 I 2 I 2 I 8 I I 3 I 3 I 3 I 3 I 3 I 3 I 3 I 3 I 3 I 8 I I 4 I 4 I 4 I 4 I 4 I 4 I 4 I 4 I 4 I 8 I I 5 I 5 I 5 I 5 I 5 I 5 I 5 I 5 I 5 I 8 I I 6 I 6 I 6 I 6 I 6 I 6 I 6 I 6 I 6 I 8 I I 9 I 9 I 9 I 9 I 9 I 9 I 9 I 0 I 9 I 8 I
20 29 10 22 19 10 21 22 19 10 31 32 19 10 31 12 23 19 10 41 22 23 19 10 31 32 13 14 19 10 51 12 33 14 19 10 51 12 23 14 15 19 10 61 22 13 14 25 19 10 51 32 13 14 15 16 19 10 71 12 23 14 25 16 19 10 61 32 13 14 15 16 17 19 10 81 12 23 14 15 26 17 19 10 71 32 13 14 15 16 17 18 19 10 91 12 23 14 15 16 27 18 19 10 81 32 13 14 15 16 17 18 29 10 81 22 23 14 15 16 17 28 19 10 71 42 13 14 15 16 17 28 19 = A 10 81 22 13 24 15 16 27 18 19 = B 10 71 42 13 14 15 16 17 28 19 = A etc...Can anyone find longer than 21 steps?
I haven't found any numbers with cycle-length longer than 3. Is there a mathematical proof that this is always so? Is there a proof that the longest is 21 terms, and that all series always terminate?
[I'd be interested in any longer cycles, or the proofs Dave mentions. - KD].
I found your article about self referential numbers in 980608 and I have some facts to supplement. To shorten things I denote with T the describing operator for integers, so T(1334) = 112314, T(22) = 22 and so on. 1) There are exactly 109 fixed-points (= self-refential numbers), i.e. n with T(n) = n. I examined this topic about a year ago and sent this sequence to Neil Sloane. They are sequence A047841, see http://www.research.att.com/cgi-bin/ access.cgi/as/njas/sequences/eisA.cgi?Anum=A047841 The reason I found your page was my search about this topic using some of these fixed-points as input for a search engine e.g. 21322314. 2) You ask if the list always ends in fixed-point or a cycle. The answer is Yes. This is a consequence from 2 facts: A) For every n with more than 21 digits T(n) < n is valid and B) For every n with not more than 21 digits, T(n) has also not more than 21 digits. As a consequence the sequence n, T(n), T(T(n)) ... enters and than remains in the bounded area from 1 to 10^21, so it must end in a cycle (see a fixed-point as a 1-cycle) To show A) and B) consider that T(n) has a maximum nuber of digits if the digits in n are distributed as equal as possible. If you have for example already 100 '2's in n you need another 900 '2's to "generate" a new digit in T(n). But if you have just 10 '2's in n you need only 90 other '2's to get a new digit in T(n). If n has k digits, T(n) reaches the maximum number of digits if eyery digit occurs at least int(k/10)-times. This leads to A) and B) 3) You ask if the number of steps until a list with a certain integer as start can have a length > 21. The answer is the list can be abitrarly large. For every a(1) > 22 not ending with a zero, there is a a(2) > a(1) with T(a(2)) = a(1) and a(2) not ending with a zero as well. Iterating this process you find a sequence of numbers a(1) < a(2) < a(3) < a(4) < .. with T(a(2)) = a(1), T(T(a(3)) = a(1), T(T(T(a(4))) = a(1) and so on. So the "pre"-phase before reaching the cycle can be abitrarly large, because the cycle doesn't start until a(1). Finding a(2), a(3), ... is easy: Let's start e.g. with a(1) = 41. Then a(2) is of course 1111. a(3) is 11111......111111 (the number consisting of 111 '1's). a(4) is the integer consisting of (10^111-1)/9 '1's and so on. 4) I don't know if there are any terminating cycles with lengths other than 1 or 3. One has "only" to examine all numbers smaller than 22 digits. But this is too much for a progam. But if you start examining all numbers beginning with 1 you can stop examining a number n if T(n) < n cause T(n) has already be checked. This reduces the nessecary run time for the program, but I think this is still too much.
Source: Unless noted, these are original.