Problems & Puzzles:
Puzzles
Puzzle 207. The
inventory
sequences and the selfinventoried numbers
For sure you know the popular look
and say infinite sequence (A005150):
1, 11, 21, 1211, 111221, 312211,
13112221, 1113213211, 31131211131221, 13211311123113112211,
11131221133112132113212221,
3113112221232112111312211312113211, ...
Let's take a closer examination of the
way of calculating each member of the sequence.
For example from 1211 you look the
first group of 1's and say '11'; then you see the next group of 2's and say
"12"; finally you see the last group of 1's and say "'21', giving the whole
next member of the sequence 111221.
But what if instead of taking
partial views of the digits of 1211 you take exhaustive views of
all the digits, as encountered, and say the total of each? In doing so, the
next member to 1211 would be 3112 instead of 111221.
Proceeding this way from a starting
number "1" this is what is obtained (A063850)
1,
11, 21, 1211, 3112, 132112, 311322, 232122, 421311, 14123113, 41141223,
24312213, 32142321, 23322114, 32232114.
Now the sequence 'ends' in a cycle of
'period 2' because from 23322114 follows 32232114 but from 32232114 follows
23322114.
Up to here we are just reviewing well
known facts about two well known sequences.
A.
Now here comes a challenging assertion from Clifford
A. Pickover. He says ^{(*)} that
this last kind of sequences were studied by certain Roger Hargrave
from West Sussex, U.K. and that:
"oddly, he
[Hargrave] believes that all his sequences
[starting in any number?] finally oscillate
between 23322114,
32232114. Can you prove this?"
Maybe a better phrase would have been "can
you prove or disapprove this"? Because the real fact is that the
Hargrave's statement is easily disapproved.
Just a couple of examples:
1) If the starting number is
22,
the sequences 'ends' immediately in a cycle of period 1.
2) If the starting number is
40,
the sequence 'ends' in a cycle of period 4:
40,
1410, 211410, 12311410, 4112131410, 2451121310, 221415411310, 225124151310,
322541141310, 232215244110, 421331152410,
242241231510, 422431131510, 242223411510, 422413311510
BTW, Hargrave named these
sequences Gleichniszahleninventar,
name that I like it because of the "inventoryaction" idea reflected in the
name.
Following this idea I will call these
sequences as "inventory sequences" and "selfinventoried
numbers" to such starting numbers that immediately end in a cycle of
period 1, as the number 22 does.
B.
Other than 22, are there more selfinventoried composite
numbers?
Yes; they are a lot of them. Here are
the first 10 of them:
Selfinventoried
composites, by C: Rivera, January, 2003 
22, 33311012, 33311014, 33311015, 33311016,
33311018, 33311019, 33311210, 33311214, 33311215, ...

C.
Are there selfinventoried primes?
Yes, but they are sparser than the
selfinventoried composites. Here are the first 10 of them that I have
gotten by an exhaustive search below 2^32:
Selfinventoried primes, by C: Rivera, January, 2003 
33311017, 33311417, 33311519, 2233311419,
2233311617, 3322311017,
3322311617,
3331102219, 3331192217, 3331221917, more?

D.
Now, let's see the inventory sequence starting with the
prime 7:
7,
17, 1117, 3117, 132117, 31131217, 23411217, 2213143117,
2241231417, 3224311317, 3322143117, 3322311417,
3322311417
In this case we notice that a) there
are 3 consecutive primes (7, 17 & 1117) b) the sequence ends in a cycle of
period 1 with a selfinventoried composite number (3322311417)
Here is a list of the starting prime
number that generates a sequence of K primes in its corresponding
inventory sequence:
K 
Inventory
sequence of K primes
(1 to 7 by CR) 
1 
2 
2 
3, 13 
3 
7, 17, 1117 
4 
11173, 311713,
233117, 12232117 
5 
29527, 22191517,
2231191517, 221341191517,
22511314191517 
6 
8986717,
2819162711, 221841191617,
22511814191617, 2215611814191617
2271152618141917 
7 
2222228557 62182517 162221181517
511632181517 25511613121817 22256116131817 32152651131817 
8 
See below a solution
by Jon Wharf. Is this the least
solution? 
And our questions are:
 Do all the
inventory sequences end in a cycle, or is there a starting number
generating an endless inventory
sequence?
 Is there a finite
number of cycles at the end of the inventory sequences? If so, what is
the cycle with the largest period?
 Can you find a
starting prime number that generates
more than
7
consecutive primes
in its corresponding inventory sequence?
 Are there
infinite selfinventoried composite
numbers? If not, can you find the
largest one of these?
 Are there
infinite selfinventoried prime
numbers? If not, can you find the
largest one of these?
 Can you extend
the list of the selfinventoried
primes given above?
 Can you find a
starting number that, in its own inventory sequence, generates an
ending cycle composed only of prime numbers?
__________
^{(*)}^{
}See p. 321,
"Wonders of Numbers", Oxford 2001
Solution:
Joseph L. Pe and Jon Wharf
sent excellent answers for this puzzle. Jon Wharf brings as a follow up of
his search new specific challenges to question 3, 4 & 5.
Joseph L. Pe wrote:
ANSWER TO Q1. Any inventory sequence ends in a
cycle.
Proof. Let D(n) denote the number of digits in the
decimal expansion of n, and let g denote the inventory sequence generator
function (for example, g(1211) = 3112). Then the inventory sequence
beginning with N, denoted IS(N), can be written as: N, g(N), g(g(N)),....
Note that if any two terms of IS(N) are equal, then IS(N) ends in a cycle.
I. If the digit d = 0, 1, ..., 9 appears exactly
once in n, then d contributes two digits to g(n), hence d increases the
number of digits by 1 in going from n to g(n). (For example, g(5) = 15 and
g(12) = 1112.) If d appears exactly twice in n, then the corresponding
increase is 0. (For example, g(33) = 23). But if d appears exactly k > 2
times in n, then d decreases the number of digits by kD(k)1 > 1. (For
example, g(2222) = 42 where k = 4 and the net decrease is 411 = 2
digits; and g(1111111111) = 101 where k = 10 and the net decrease is
1021=7 digits.)
II. Lemma. If D(n) <= 20, then D(g(n)) <= 20.
For suppose that n has at most 20 digits. The digits
of n can be partitioned into two finite sequences: S, consisting of the
first occurrences of digits of n, and S', consisting of the other
occurrences of the digits. (For example, if n = 11133245, then S =
{1,3,2,4,5} and S' = {1,1,3}.) By I above, each member of S increases the
number of digits by 1 in going from n to g(n), whereas each member of S'
either cancels a potential increase of 1 (when the digit occurs exactly
twice in n) or decreases the number by at least 1 (when the digit occurs
more than twice in n). But S has at most 10 members, so there are at most
10 increases in the number of digits from n to g(n). Hence, in the case
that 1 <= D(n) <= 10, clearly D(g(n)) <= 20. In the case that 11 <= D(n)
<= 20, then S' has at least D(n)  10 elements, and the net increase can
be at most 10  (D(n)  10) = 20  D(n), so that, D(g(n)) <= D(n) + (20 
D(n)) = 20. In any case, if D(n) <= 20, then D(g(n)) <= 20.
III. Lemma. If D(n) > 20, then g(n) < n.
For using the notation of II above, S can have at
most 10 terms, whereas S' must have at least 11 terms, one of which must
decrease the number of digits going from n to g(n) (since there can be no
more than 10 of the digits that occur exactly twice). That is, S has fewer
terms than S'. Hence, g(n) will have fewer digits than n, implying g(n) <
n.
IV. By repeatedly applying III above, if necessary,
one obtains a term M of IS(N) such that M <= 10^20 (that is, D(M) <= 20).
By II, all terms of IS(N) from M onwards are <= 10^20. Since these
infinitely many terms can take only 10^20 possible values, at least two of
them must be equal. Therefore, as noted above, IS(N) must end in a cycle.
Note: 10^20 is probably the sharpest upper bound for cycle terms. For
example, if
N =
12345678901234543212456778605948376273848494934857594839304384/
000000009303030, then IS(n) ends in the cycle 24228118271513161910,
42142871171513161910, 24228118271513161910, .... which are 20digit
numbers.
ANSWER TO Q2.
I. There are only a finite number of cycles at the
end of inventory sequences.
Proof. Since all but finitely many terms of IS(N)
are <= 10^20, all the terms of a cycle must be <= 10^20. But each cycle is
just a finite sequence of integers which are all distinct. Hence, the
number of possible cycles <= the number of permutations of the set {1, 2,
..., 10^20} = (10^20)!, which is a finite number.
II. Of course, this is an absurdly large upper
bound. Probably the largest period is 4, first attained at N = 40: 40,
1410, 211410, 12311410, 4112131410, 2451121310, 221415411310,
225124151310, 322541141310, 232215244110, 421331152410, 242241231510,
422431131510, 242223411510, 422413311510, 242241231510, .... I have
checked that no number < 10^6 has a greater period.
In principle, it would be simple to prove that no
inventory sequence has a cycle length > 4. By the previous proof, any
cycle begins with a number <= 10^20. Hence, the claim would be proved by
just checking that IS(N) has cycle length <= 4, for all N <= 10^20. I
leave the verification up to someone with sufficient computing resources.
Q4 and Q5: There can be only finitely many
selfinventoried numbers (hence, only finitely many selfinventoried prime
numbers and composite numbers.) This is because, by the lemma III I proved
below, any n > 10^20 must have g(n) > n; so if n is selfinventoried,
then we must have n <= 10^20
[This is slightly incorrect, see
Jon Wharf's solution to the question
Q4.]
***
Jon Wharf wrote:
I've looked at that first 5 questions and the short
version is:
1) all the inventory sequences end in a cycle
2) there is a finite number of cycles, largest
period appears to be 4 (not definite)
3)
222222222222222222222288577755557777777777777777777 (which is probably
prime) leads to seven more primes, eight in all
4) Finite selfinv composites; my largest is
221111918171615141310.
5) Finite selfinv primes; my largest is
1111918171614101513.
...
Define the inventory function inv(n) as the
inventory number generator.
An inventory number is one that can be generated as
an inventory for another number (a result of inv(n)). Almost all numbers
are inventory numbers; the requirements are:
a) at least two digits and the last digit not zero
> inventory
b) at least four digits, ends in zero, and contains
two consecutive nonzero digits excluding the leading digit > inventory
c) the number ten > inventory (inv(0))
Otherwise it is not an inventory number, eg 10000,
120304050. Interestingly this means all primes > 10 are inventory numbers.
Define a stable inventory number m as an inventory
number whose own inventory number inv(m) has at least as many digits as m.
Clearly any terminal cycle sequence must include a stable inventory number
(at least every other term, in fact).
Now an inventory number consists of at most ten qd
sets, where q is quantity and d is digit. Sum of q<i> is the number of
digits in the original number. In order for an inventory number k to have
100 digits it must inventorise a number of at least 10^9 (=10*10^8)
digits, and inv(k) will have at most 30 digits (=(2+1)*10), inv(inv(k))
will have at most 22 digits (=3+3+2*8), and inv(inv(inv(k))) can only have
at most 21 digits. So it is clear that a stable inventory number has 21 or
less digits. It's also clear that any 21digit number will inventorise to
21 or less digits, so this is a oneway limit  no inventories 21 digits
or less will produce inventories over 21 digits. Also, a sequence starting
in nonstable numbers will quickly move to include stable numbers.
This gives us directly that (Q1) the number of
stable inventory numbers is finite and so there are no endless inventory
sequences, (Q2) there are a finite number of end cycles, and (Q4 & 5)
there are only finite selfinventoried composites and primes.
(Q2) Looking at the ending cycle of any sequence, it
must have a constant number of qd sets, since once a particular d occurs
it never leaves. Some of these will be "active" and some "passive" 
"active" meaning that the q changes for this d in the cycle and "passive"
meaning it never does. All active qd sets occur before any passive qd
set, in the end cycle. There is obviously an opportunity to reorder the
passive sets at the end to produce primes.
For stable 21 digits, one q<i> must be 10 or more,
so there will be at least 7 q<i> == 1. So it must be q<1> >= 10. The
largest other possible q value would be 3 (10+3+8*1=21). The possibilities
for nonunit q are {10,3}, {10,2,2}, {11,2} and {12}. {10,3} > {10,2,2}
is nonstable > 20 digits > {8,2,2,2}. {12} > {11,2} becomes
selfinventorising.
For 20 digits, there are more possibilities for
nonunit q: {11}, {10,2}*, {9,3}*, {9,2,2}*, {8,4}*, {8,3,2}*, {8,2,2,2}*,
{7,5}*, {7,4,2}*, {7,3,3}*, {7,3,2,2}#, {7,2,2,2,2}*, {6,6}*, {6,5,2}*,
{6,4,3}*, {6,4,2,2}#, {6,3,3,2}#, {6,3,2,2,2}#, {6,2,2,2,2,2}*, {5,5,3}*,
{5,5,2,2}*, {5,4,3,2}*, {5,4,2,2,2}#, {5,3,3,3}*, {5,3,3,2,2}#,
{5,3,2,2,2,2}*, {5,2,2,2,2,2,2}*, {4,4,4,2}*, {4,4,3,3}*, {4,4,3,2,2}#,
{4,4,2,2,2,2}*, {4,3,3,3,2}#, {4,3,3,2,2,2}*, {4,3,2,2,2,2,2}#,
{4,2,2,2,2,2,2,2}#, {3,3,3,3,3}*, {3,3,3,3,2,2}*, {3,3,3,2,2,2,2}*,
{3,3,2,2,2,2,2,2}*, {3,2,2,2,2,2,2,2,2}* and {2,2,2,2,2,2,2,2,2,2}.
({7,4,2} and {8,2,2,2}) form a 2cycle reached by all *, {7,3,2,2} self
2cycles, reached by all #, and both {11} and {all 2} get to 21 digits and
{11,2}. None are selfinventorised.
For 19 digits, only the set missing 2, {11}\2 is
stable, and selfinventorises.
18 digits strongly resembles 20 digits, with end
2cycles
({6,4,2},{7,2,2,2}) and ({6,3,2,2})
17 and lower odd numbers of digits cannot give
endcycles.
And so on. Once the number of ones starts to
interact with the number of twos or threes we stand some chance of more
interesting behavior, but I haven't yet found an endcycle of more than 4.
(Q3) No doubt there is a prime which inventorises to
2222228557 or one of the similar primes, eg 2222855227 (22:2, 2:8, 5:5,
22:7, should be plenty there and "only" 51 digits).
222222222222222222222288577755557777777777777777777 seems to
work. (thanks to Joe Crump's ZZMath Excel addin)
(Q4) The largest (21digit) selfinventoried number
I've thought of is 221111918171615141310.
This and all scrambling are divisible by 3. Eleven ones, two twos and one
of each other digit is the only possible selfinventoried solution set for
21 digits so primes will be smaller.
(Q5) There are no 20 digit selfinventoried numbers
so we drop to the {11}\2 solution for 19 digits and
1111918171614101513 is my candidate for
largest selfinventoried prime.
***
And the added challenges by J. Wharf
are: can you improve his findings for the questions
3, 4 & 5 (numbers in boldred)?
***
