Need Help Understanding a Pattern I Found

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The discussion centers on a mathematical pattern involving consecutive integers where the sums of three consecutive numbers yield results that are multiples of 6. The pattern is established through the formula (a-1) + a + (a+1) = 3a, demonstrating that if 'a' is odd, the resulting sums are even and divisible by 6. The conversation also touches on the continuity of the odd-even pattern in the sequences presented and suggests exploring palindromic numbers as a related topic for further study.

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  • Understanding of basic arithmetic operations and properties of odd and even numbers
  • Familiarity with algebraic expressions and simplification techniques
  • Knowledge of number theory concepts, particularly divisibility rules
  • Basic programming skills, particularly in Fortran for numerical exploration
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  • Research palindromic numbers and their properties
  • Explore the concept of divisibility in number theory, focusing on multiples of 6
  • Learn about arithmetic sequences and their applications in mathematics
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Emanresu56
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I'm definitely sure this has already been discovered by some mathematician way back, but I just discovered it today.

Here it is:

1+2+3=6
3+4+5=12
5+6+7=18
7+8+9=24

Etc.

What is this called (if it's called anything), and how does it work? And I'm mystified as to how the first numbers in each equation are odd (1, 3, 5, 7), and then there are even numbers (2, 4, 6, 8), and then there are odd numbers (3, 5, 7, 9), and then there are even numbers again (6, 12, 18, 24). Maybe I just inadvertently set it up that way?

Thanks! :biggrin:
 
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In general, you have

(a-1) + a + (a+1)

simply collect the 1's to get

a + a + a = 3a

So of course, if a is an even number, 3a will be an even number.
 
You "set it up that way" when you chose the first number to be 1, an odd number. If n is an odd number, say, n= 2k+ 1, then the next number, n+ 1= 2k+1+1= 2k+ 2= 2(k+ 1) is even, and the last, n+2= 2k+1+ 2= 2k+ 2+ 1= 2(k+1)+ 1 is again odd. And by taking the last number on one line as the first number on the second line you have guarenteed that "odd, even, odd" pattern continues.

As for the last column, for n odd, n= 2k+1, n+ (n+1)+ (n+2)= 2k+1+ (2k+2)+ (2k+3)= 6k+ 6= 6(k+1) so the last column is not just even but is always divisible by 6.
 
Perhaps a good direction to take your study is into palandromic numbers such as 11,22,33... then 111, 121, 131, ... then 1221, 1331, 1441 and so on. I did wrtie a small fortran routine years ago. but may work better under the control of string manipulation than arithetic.
 

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