Problems & Puzzles: Puzzles

Puzzle 107. K Consecutive Smith numbers

A Smith number is such that the sum of its digits is the sum of the digits of all its prime factors

Example: 666 = 2*3*3*37 & 6+6+6 = 2+3+3+(3+7 ). 

But, are there consecutive Smith-numbers? In which extent?

I have calculated the least K-tuplets of consecutive Smith numbers for K=1, 2, 3, 4 & 5.

K

First member of the least K-tuplet

1

4

2

728

3

73615

4

2561678

5

4463535

6

?

Questions:

1. Maybe you would like to extend the sequence
2. Can you argument if are there this kind of numbers consecutive in any extent (K)?


Solution

Shyam Sunder Gupta wrote the 7/2/2001:

a) Consecutive Smiths mentioned for K= 4 & K=5 is incorrect , if primes are not considered as Smiths. Because for K= 4 , the numbers 2561678 ,2561679 & 2561680 are Smith numbers but 2561681 is prime. Similarly for K= 5 the numbers 4463535,4463536,4463537 & 4463538 are Smith numbers but 4463539 is prime.

b) If primes are considered as Smiths then for K=1, the smith number 4 is incorrect. because 2,3,5 are primes so smiths and hence 2,3,4,5 are four consecutive Smith numbers.

I feel that the purpose is to find consecutive non-prime Smith numbers. I have done a Systematic study of all Smith numbers up to 10^8 after writing code in Fortran and the correct results are as follows. 

K  Starting Smith Number
1  4 
2  728
3  73615
4  4463535
5  15966114

In fact there are only two sets of 5- consecutive numbers starting from 15966114 (as mentioned above) & 75457380 up to 10^8 . A further study reveals that up to 4*10^8, there are no 6-consecutive Smith numbers and there are only 6 , 5-consecutive Smith numbers including those mentioned above.

***

More on this issue in Puzzle 247

***


Records   |  Conjectures  |  Problems  |  Puzzles