Problems & Puzzles:
Puzzles
Puzzle 251.
Pointer primes
Joseph L. Pe proposes the
following puzzle:
The product of the digits of the
prime number 23 is 2 x 3 = 6. Note that 23 + 6 = 29, the next prime
number. The prime 23 can be said to "point" to the next prime number 29
in the sense that when 23 is added to the product of its digits, 29 is
obtained.
We define the prime number p to be
a pointer prime if the next prime after p can be obtained from p
by adding to p the product of the digits of p.
The pointer primes not exceeding 10^{8}
are:
23, 61, 1123, 1231, 1321, 2111,
2131, 11261, 11621, 12113, 13121, 15121,
19121, 21911, 22511, 27211, 61211, 116113, 131231, 312161, 611113,
1111211, 1111213, 1111361, 1112611, 1123151, 1411411, 1612111, 2111411,
2121131, 3112111, 3116111, 3221111, 4121113, 9111341, 11113321,
11132123, 11221211, 11264111, 11281211, 11314111, 11316211, 11361211,
11413111, 11611121, 11922121, 12111163, 12111313, 12111331, 12111511,
12122111, 13121117, 14111131, 14111311, 19122211, 21182111, 21212111,
23311111, 27212111, 41122111, 41211221, 61114211, 81111211, 111111163,
111131213, 111219211, 111221221, 111316111, 111611113, 112121213,
112212211, 114131141, 115121221, 115121311, 116112131, 119112113,
121115311, 121122151, 121151341, 121235111, 131121131, 131211181,
131261111, 136211111, 152211121, 163111111, ...
Note that 23 is the only number in
the list that does not contain the digit 1.
Question. Are
there other pointer primes that do not contain the digit 1?
Solution:
Contributions to the only question
came from Jon Wharf, J Van Delden and Faride Firoozbakht:
Wharf wrote:
Question. Are there other pointer
primes that do not contain the digit 1?
No.
Examination of the maximal prime
jumps at http://www.trnicely.net/gaps/gaplist.html shows that the gaps
increase much more slowly than the minimum digitproduct (MDP) without
ones (MDP2+) for the bounding primes.
For example, 3 maximal jumps are
encounter in 8digit numbers:
180 after 17051707
210 after 20831323
220 after 47326693
but the minimum digit product is
3*2^7 = 384.
The ratio between these numbers
continues increasing until the largest confirmed maximal jump of 1198
after 55350776431903243 compared to the corresponding MDP2+ of 196608
for 17 digits (ratio>164).
It is clear that the maximal prime
jumps are increasing at a rate consistent with the density of primes
whereas the MDP2+ increases much faster. There is no way back and no
more oneless pointer primes.
***
Van Delden wrote:
A dirty heuristic argument:
A quick search on Mathworld revealed that the
maximum Gap between successive primes less than n would be asymptotically
ln(n)^2. (Cramer's conjecture).
In order for primes to be successive we can't have a
0 as a digit. Since in the proposed question the 1 is prohibited as well,
every digit is at least 2. (And at least 1 is bigger than 2, otherwise we
can't have a pointerprime).
Starting with n consisting of d 2's (gives smallest
gap, not prime though, but simplifies the argument).
Our Gap would then be: 2^d=(ln(n))^2. Which leads to
n asymptotically exp(2^(d1)). But n itself would be about 2*10^(d1).
Solving these two expressions for equality gives two solutions for d, d
being about 1 and 4. If d>4 the 'required size of n' (by Cramer's
conjecture) will be far greater than the actual size: d=8 digits would
give Cramer's n about exp(2^7)  3.9*10^55, quite a bit bigger!
So I would say no. To prove there are no pointer
primes of the required form, would entail using an upper bound for the
maximum prime Gap, finding bounds for the solution(s) of a similar
more precise equation for n, giving a range to search for pointer primes p
less then (say) N. (Probably done already).
***
Faride wrote:
I have proved (in my thesis in 1991) that If
conjecture 30 is true then for each n d(n) < 3*Log(prime(n))^2 (**)
(and d(n)<(1+epsilon)*Log(prime(n))^2 for
sufficiently large n)
where d(n)=prime(n+1)prime(n).
Now suppose that p=prime(m) is a pointer prime and
10^(n1) < p < 10^n so p is a ndigit prime if p doesn't contain the digit
1 then d(m) >= 3*2^(n1),and if conjecture 30 is true d(m) <
3*Log(prime(m))^2
So 3*2^(n1) < 3*Log(p)^2 and since p < 10^n we have
2^(n1) < Log(10)^2*n^2 ,but this relation is true only for n=1,2,...,10
So if the conjecture is true there is no pointer prime greater than 10^10
which doesn't contain the digit 1. So the numbers greater than 10^8 with
one digit 3 must check are 222222223, 2222222223,but they aren't prime.
If p contains at least one digit not equal to 2 and
3 then by using (**) p must be less than 10^8 . So if the conjecture is
true there is no other pointer prime that does not contain the digit 1.
***
