Need Help Understanding a Pattern I Found

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The discovered pattern involves summing three consecutive integers, resulting in a formula of 3a, where a is the first integer. This leads to the conclusion that if a is odd, the sums will always yield even results, specifically multiples of 6. The sequence of numbers alternates between odd and even, creating a consistent pattern that continues with each new set of three integers. The discussion suggests exploring related concepts like palindromic numbers for further mathematical inquiry. Overall, the pattern highlights interesting properties of odd and even integers in arithmetic sequences.
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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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...