I’ve also discovered that if you take the d, +1 and divided by 2, you find the square root. So if 64 has a d of 15, 15+1=16 and 16/2= 8. Also, any d with 9 as second value (d=9 or d=19) is a multiple of 5. Therefore if you have something like 9025, you can x the difference between the difference and add 50 and you find 189d is = to 9025. Therefore, 189+1/2 = 95.

The equation is simplified to sqr=d+1/2

# Surprising Patterns in the Square Numbers (1, 4, 9, 16…)

**samuela**#101

**ghee**#102

Can you help me with this? There is a pattern between these numbers but I can’t seem to find it.

1,6

2,9

3,12

4,16

5,18

6,21

7,24

**jk0**#103

@Cassidy

There’s no immediate pattern unless 4, should in fact be 15.

Then it would simply be n*3+3 or (n+1)*3.

**loi**#108

what’s the general formula of this sequence and what type of sequence is this 3,6,7,-2. . . . . .?

**hyperthreading**#109

@Sol Lederman (post #15 … is almost 5 years old, but I saw it today … ) wrote: "… A great follow up article would be to show how the sums of cubes is related to the volume of a pyramid.“

Maybe he meant: “… sum of squares …”, not “… the sum of cubes …”, because sum of squares (1 to n) is:

n*(n+1)*(2n+1)/6, can be written as n*(n+1)*(n+1/2)/3 or n^3/3 + n^2/2 + n/6, so we may deduce that the volume of {cone or pyramid} might be: area_of_the_base * height /3 , and the sum of cubes is: (n*(n+1)/2)^2 =

(sum_of_numbers_1_to_n)^2, and this would be a sqare of some (“triangular_like_shape”) area, which is a

"hypervolume” of a 4_dimensional "object "…

**hyperthreading**#110

There are a lot of “problems above”, be more specific. For the sum of “n-th” powers of numbers 1 to “n” you do not need any program. I had learned that long before I learned to program in Basic, FORTRAN, Pascal, C, COBOL, PL/I … while Object Oriented Programming and C++ existed, but at universities …

**jeremy**#111

@Chris Nash

[comment #28, #29, #98]

The “trick” still holds for 3^10 being divisible by 3 and therefore (3^10)^2 = 3^20 being divisible by 9.

The idea is that the sum of the digits of 3^10 (=59,049 -> 5+9+0+4+9=27) is divisible by 3, while the sum of the digits of 3^20 (=3,486,784,401 -> 3+4+8+6+7+8+4+4+0+1=45) is divisible by 9.

Note that 3^10 was divisible by 9 in the first place, but even if it was only divisible by 3 (and not by 9), its square would be divisible by 9.

**sidzz**#112

… so interesting… Amazed that we can play with square numbers… Soon i will find a new pattern and post. M working on that.Till then keep waiting…

**chrisnash**#113

@Sean

The thing about “as long as the number being squared is divisible by 3, that the sum of the numbers in the answer of the square is equal to 9” isn’t true for (3^20) even though that number is divisible by 3 and it’s square root is a whole number. The digits of 59049 don’t add up to 9!

**johnding**#116

I’ve seen this before, commenting on the “2n+1” beginning part. This is pretty obvious because 2n+1 would be the difference from n^2 and (n+1)^2 so n^2+2n+1 would be the next square. You could’ve done (n+1)^2 and just got those values, which equates to a quadratic function.

**Matt_Clopton**#119

The first 20 square numbers are obviously

1

4

9

16

25

36

49

64

81

100

121

144

169

196

225

256

289

324

361

400

The last digets repeat 0,1,4,9,6,5,4,1,0 as you probably already know. In between squares 4-7 the first digits increase by 1 and in squares 8-13 the middle number goes up by 2, 14-17 the middle numbers go up be 3. And so on every section skips 3 then 5.