Harmonic sequence puzzle

 
 
Definition:   A harmonic sequence is a sequence whose reciprocals form an arithmetic sequence.

 
 

1189   has 4 divisors:     1   29   41   1189

1189 \; = \; 29 \times 41

1189 \; + \; 1 \; = \; 1190

29 \; + \; 41 \; = \; 70

1190 \; = \; 17\times (29 + 41) \; = \; (17\times 29) \; + \; (17\times 41) \; = \; 493 + 697

493 \; = \; 17\times 29 ………. 697 \; = \; 17\times 41

 

(493, \; 697, \; 1189)   form a harmonic sequence since

1/493 \; - \; 1/697 \; = \; 1/697 \; - \; 1/1189 \; = \; 12/20213

 

determine a method to find sets of positive integers   (a, b, c)   such that

a,   b,   c   forms a harmonic sequence
a   <   b   <   c
a   +   b   =   c   +   1

 
Then, do the same for sets of positive integers   (a, b, c, d)   such that

a,   b,   c   d   forms a harmonic sequence
a   <   b   <   c   <   d
a   +   b   +   c   =   d   +   1

 
                                Part 1         
 

Solved by pipo

a,   b,   c
——————-
(4, 3, 6)
(21, 15, 35)
(120, 85, 204)
(697, 493, 1189)
(4060, 2871, 6930)

Recursion for the c:   c(n) = 6*c(n-1) – c(n-2)

 
                                Part 2         

 
The lowest integer that has 4 factors that are adjacent is 12:

12   has 6 divisors:    1   2   3   4   6   12

1/12,   2/12,   3/12,   4/12   in arithmetic progression

the reciprocals are   12,   6,   4,   and   3   respectively.

(3, 6, 12)   in arithmetic progression if we add 9
(3, 6, 12)   in geometric progression if we add 24

 
Set ……………. Arithmetic ….. Geometric ….. Harmonic
3,4,5, 6,12,24 ….. 3, 4, 5, 6 ….. 3,6,12,24 ….. 3,4,6,12
3,4,6, 9,12,24 ….. 3, 6, 9,12 ….. 3,6,12,24 ….. 3,4,6,12
3,4,6,12,18,24 ….. 6,12,18,24 ….. 3,6,12,24 ….. 3,4,6,12

1/3   –   1/4   =   1/4   –   1/6   =   1/6   –   1/12   =   1/12
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

2 Responses to Harmonic sequence puzzle

  1. pipo says:

    Here are some solutions for the first puzzle:
    (a, b, c) =
    (4, 3, 6), so 1/3 – 1/4 = 1/4 – 1/6 and 4 + 3 = 6 + 1
    (21, 15, 35)
    (120, 85, 204)
    (697, 493, 1189)
    (4060, 2871, 6930)

    Recursion for the c: c(n) = 6*c(n-1) – c(n-2)

    pipo

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