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September 25th, 2019, 10:57 AM   #1
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Question (!MUST READ!) So, I found this odd sequence...

I was fiddling around with some square numbers two days ago, when I started doing this (finding differences):

0,1,4,9,16,25,36,49,64,81,100,121,144,169
1,3,5,7, 9 ,11,13, and you see where this is going. All the numbers in this row are odd numbers, two different.

What about cubed...

0, 1 , 8 , 27,64,125,216,343,...
1, 7 ,19,37 ,61, 91 ,127,... (continuing)
6,12,18,24 ,30, 36 , and now we have six different.

Now here, I went to the 4th power:

0 , 1 , 16 , 81 ,256,625,1296,2401,...
1 ,15, 65 ,175,369,671,1105,...
14,50,110,194,302,434,...
36,60, 84 ,108,132,...
24,24, 24 , 24 , and now we have 24.

0th power does this:

0,0,0,0,0,...
0,0,0,0,... (you get it)

1st power is this:

0,1,2,3,4,5,6,...
1,1,1,1,1,1, and now 1s.

Next, I arranged them like this:

0,1,2,6,24,...

And before I go on, I would like to note something. From one onward, if you multiply the 1 by two, you get the next object in the sequence, 2. 2*3=6, 6*4=24. I have confirmed this up until the multiplication by seven. So why is 0*1=1?

Anyway, proceeding:

0 , 1 , 2 , 6 , 24 ,120,720,...
1 , 1 , 4 , 18 , 96 ,600,...
0 , 3 , 14 , 78 ,504,...
3 , 11 , 64 ,426,...
8 , 53 ,362,...
45 ,309,...
264,...

Now, this may be infinite, but, using the information that you gleaned from this, I would like an equation, a possible answer to the 0,1,2,6,24... sequence, and I would like to know a way to submit this as an unsolved problem in math.

This may be unsolvable, just to let you know. It appears that the multiplication rate is higher than the deduction rate. But, this may work against itself. And for right now, I would be happy to leave the answer at, of course, 42.
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September 25th, 2019, 08:02 PM   #2
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https://oeis.org/search?q=0+%2C+1+%2...lish&go=Search
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September 25th, 2019, 08:29 PM   #3
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The sequence you discovered is the factorial sequence. By definition, $n! = n(n-1)(n-2)\dots (2)(1)$ and by definition 0! = 1. There is a good reason for this but its besides the point at present. So the first few factorials are: $0! = 1, 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720$, etc etc. See the following link for more details.
https://en.wikipedia.org/wiki/Factorial

Now, the reason they appear in your construction is the following. The first is that consecutive integer powers like you are computing are given by the binomial theorem as
\[(m+1)^k - m^k = \sum_{k = 0}^n \binom{n}{k} m^k - m^n = \sum_{k = 0}^{n-1} \binom{n}{k} m^k \]
Notice that this is a polynomial expression in $m$. Then the divided difference formula for polynomial interpolates guarantees that after taking divided differences $n$ times, the resulting differences will become constant. See the following:
https://en.wikipedia.org/wiki/Newton_polynomial

Now, why should that constant be exactly $n!$? That is because these divided differences end up giving multiples of the coefficients of the derivative of the polynomial. The last divided difference which results in those constant terms is the result of taking $n-1$ derivatives of a polynomial of degree $n-1$. The leading term has the form $\binom{n}{1}m^{n-1} = n m^{n-1}$ and notice it is the only term which contributes to this derivative. So after taking $n-1$ derivatives we have
\[ \frac{d^{n-1}}{dx^{n-1}}n m^{n-1} = n(n-1)! = n!\] which is the constant appearing in your divided differences.
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September 26th, 2019, 03:24 AM   #4
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oeis.org has almost all possible integer sequences.
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September 26th, 2019, 05:25 PM   #5
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Quote:
Originally Posted by tahirimanov19 View Post
oeis.org has almost all possible integer sequences.
All but finitely many, right??
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September 27th, 2019, 07:14 AM   #6
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Quote:
Originally Posted by The Chaz View Post
All but finitely many, right??
Welcome back!
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September 27th, 2019, 08:26 AM   #7
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Quote:
Originally Posted by The Chaz View Post
All but finitely many, right??
I have a slightly different understanding of right or wrong.
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September 30th, 2019, 06:32 AM   #8
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Thanks, but i don't have that understanding. English please?
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September 30th, 2019, 07:16 AM   #9
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Quote:
Originally Posted by Masterisk View Post
0,1,4,9,16,25,36,49,64,81,100,121,144,169
1,3,5,7, 9 ,11,13, and you see where this is going. All the numbers in this row are odd numbers, two different.
The difference between consecutive squares is $(n_1+1)^2 - n_1^2 = (n_1^2 + 2n_1 + 1) - n_1^2 = 2n_1 + 1$ The difference between consecutive terms of this sequence is $\big(2(n_2 + 1) + 1\big) - \big(2n_2 + 1\big) = 2$.

Quote:
Originally Posted by Masterisk View Post
What about cubed...

0, 1 , 8 , 27,64,125,216,343,...
1, 7 ,19,37 ,61, 91 ,127,... (continuing)
6,12,18,24 ,30, 36 , and now we have six different.
The difference between consecutive cubes is $(n_1+1)^3 - n_1^3 = 3n_1^2 + 3n_1 + 1$ and the difference between consecutive terms of this sequence is $\big(3(n_2 + 1)^2 + 3(n_2+1) + 1\big) - \big(3n_2^2 + 3n_2 + 1\big) = 6n_2 + 6$. The difference between consecutive terms of this sequence is $6$.

This technique generalises for any polynomial expression and can be used to determine a polynomial that generates any given finite sequence of numbers.

Reference: https://www.purplemath.com/modules/nextnumb.htm

None of this is particularly relevant to your final sequence of course, because that isn't generated by a polynomial.
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Last edited by skipjack; October 1st, 2019 at 03:02 PM.
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September 30th, 2019, 11:33 AM   #10
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Originally Posted by v8archie View Post
None of this is particularly relevant to your final sequence of course, because that isn't generated by a polynomial.
Actually, having re-read, I now see where your final sequence comes from and you will see that SDK's post contains the information that joins the dots.
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