This is the sequence of numbers n such that 3 times the reversal of n plus 1 is the number itself. In other words, n = 3*reversal(n)+1. For example, 742 = 3*247+1. In fact, 742 is the smallest number with this property. How does this sequence continue, and why?

I figured out that 783742162 is another number in the sequence. So is 783783742162162. In fact, you can keep tacking 783’s at the start and 162’s at the end, and thus get infinitely many such numbers.

Are those (with 742) all the numbers in the sequence? I don’t know, but I’m guessing yes, since otherwise you probably wouldn’t have asked this question. ðŸ™‚

The number in the largest place must be 7 and the number in the ones place must be 2.

The number in the ones place after product with 3
3 x 1 -> 3
3 x 2 -> 6
3 x 3 -> 9
3 x 4 -> 2
3 x 5 -> 5
3 x 6 -> 8
3 x 7 -> 1
3 x 8 -> 4
3 x 9 -> 7

Since n-1 = 3*reversal(n)

If the number in the ones place of n is 2, the number in the ones place of n-1 is 1, which is the number in the ones place of 7 multiplying by 3. Then 7 is the number in the largest place of n.

Let n be a 4-digit number, there are the following combinations n and its reversal
7xx2 2xx7
6xx3 3xx6
1xx4 4xx1
2xx5 5xx2
5xx6 6xx5
2xx7 7xx2
9xx8 8xx9
6xx9 9xx6

Since n > reversal(n) and (n-1)/reversal(n) =3, there’s only one possibility for the smallest and largest places combination, that is
7xx2 2xx7

Solution for the 4 digit number
Let a be the number in tens place of n, and b the number of hundreds place.
n-1 = 3*reversal(n) is converted to the flowing equation
7000+100a+10b+2-1= 3*(2000+100b+10a+7)
980+70a = 290b

No integer pairs are found for a and b. So no solution for 4 digit numbers.

## Joseph:

I figured out that 783742162 is another number in the sequence. So is 783783742162162. In fact, you can keep tacking 783’s at the start and 162’s at the end, and thus get infinitely many such numbers.

Are those (with 742) all the numbers in the sequence? I don’t know, but I’m guessing yes, since otherwise you probably wouldn’t have asked this question. ðŸ™‚

19 July 2021, 10:12 pm## Piotr:

It seems that the answer is all numbers A that can be generated (recursively) by the following productions:

A -> 742 | 742B742 | 783A162

B -> 5 | 162B783 | 5A5

So the first few are:

742

22 July 2021, 7:56 pm7425742

783742162

74257425742

7421625783742

7837425742162

742574257425742

783783742162162

74216257425783742

74257837421625742

78374257425742162

## Uniteasy:

The number in the largest place must be 7 and the number in the ones place must be 2.

The number in the ones place after product with 3

3 x 1 -> 3

3 x 2 -> 6

3 x 3 -> 9

3 x 4 -> 2

3 x 5 -> 5

3 x 6 -> 8

3 x 7 -> 1

3 x 8 -> 4

3 x 9 -> 7

Since n-1 = 3*reversal(n)

If the number in the ones place of n is 2, the number in the ones place of n-1 is 1, which is the number in the ones place of 7 multiplying by 3. Then 7 is the number in the largest place of n.

Let n be a 4-digit number, there are the following combinations n and its reversal

7xx2 2xx7

6xx3 3xx6

1xx4 4xx1

2xx5 5xx2

5xx6 6xx5

2xx7 7xx2

9xx8 8xx9

6xx9 9xx6

Since n > reversal(n) and (n-1)/reversal(n) =3, there’s only one possibility for the smallest and largest places combination, that is

7xx2 2xx7

Solution for the 4 digit number

Let a be the number in tens place of n, and b the number of hundreds place.

n-1 = 3*reversal(n) is converted to the flowing equation

7000+100a+10b+2-1= 3*(2000+100b+10a+7)

980+70a = 290b

No integer pairs are found for a and b. So no solution for 4 digit numbers.

6 August 2021, 11:07 am