Pointer Primes and Beyond

I need to acknowledge that the number associated with my diurnal age today, 27211, is the last pointer prime that I will encounter in my current lifetime, the next being 61211. So what is a pointer prime?

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 \(p\). Source.

The pointer primes up to one million are as follows(permalink):

23, 61, 1123, 1231, 1321, 2111, 2131, 11261, 11621, 12113, 13121, 15121, 19121, 21911, 22511, 27211, 61211, 116113, 131231, 312161, 611113

The conjecture is that 23 is the only pointer prime that does not contain a 1 and this seems likely to be the case. In the case of my diurnal age number we have 27211 + 28 = 27239.

It's easy enough to extend the idea of pointer primes to pointer semiprimes. In this case we can say that:

We define the semiprime number \(n\) to be a pointer semiprime if the next semiprime after \(n\) can be obtained from \(n\) by adding to \(n\) the product of the digits of \(n\).

In the range up to one million, the only such pointer semiprimes are 22, 123, 1411, 11521, and 111161 as shown below (permalink with the algorithm only considering semiprimes with distinct prime factors):

22 = 2 * 11 --> 26 = 2 * 13

123 = 3 * 41  --> 129 = 3 * 43

1411 = 17 * 83  --> 1415 = 5 * 283

11521 = 41 * 281  --> 11531 = 13 * 887

111161 = 89 * 1249  --> 111167 = 7 * 15881

It can be seen that all except 22 contain the digit 1 as was the case with 23 and the other pointer primes.

So let's go on to pointer sphenics in which case we can say that:

We define the sphenic number \(n\) to be a pointer sphenic if the next sphenic after \(n\) can be obtained from \(n\) by adding to \(n\) the product of the digits of \(n\).

In the range up to one million, the only such pointer sphenics are 222, 21122, 31114, 111122, 111711, 125111 and 912111 as shown below (permalink):

222 = 2 * 3 * 37 --> 230 = 2 * 5 * 23

21122 = 2 * 59 * 179 --> 21130 = 2 * 5 * 2113

31114 = 2 * 47 * 331 --> 31126 = 2 * 79 * 197

111122 = 2 * 11 * 5051 --> 111126 = 2 * 3 * 18521

111711 = 3 * 23 * 1619 --> 111718 = 2 * 83 * 673

125111 = 7 * 61 * 293 --> 125121 = 3 * 179 * 233

912111 = 3 * 23 * 13219 --> 912129 = 3 * 47 * 6469

Once again it is only 222 that does not contain the digit 1 while all others do. Of course, we could change the definition of a pointer prime so that the prime needs only to point to any higher prime not just the next higher prime. We need to exclude primes containing zero in this definition because the product of digits is then zero and adding zero to the original prime leaves us with the original prime! By this definition, the pointer primes would be far more frequent. For example, there would be 899 of them in the range up to 40000 (permalink).