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February 11th, 2017, 09:48 PM   #1
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Lightbulb Summing a set past the end of the set.

The sum from n=1 to k+d of a_n is equal to a_k.
We can deduce that a_n = a_(n-d) - a_(n-d-1) and if a_1 = 1 and d = 1 it results in:

[1,0,-1,-1,0,1,1,0,-1,-1,0,1,1,0,-1,-1,0,...]

When we change the value of d, it becomes more chaotic. When d=4, we get:

[1,0,0,0,0,-1,0,0,0,-1,1,0,0,-1,2,1,0,-1,3,-3,1,-1,4,-6,4,-2,5,-10,10,-6,7,...]

Some of you who are keen may notice Pascal's triangle hidden in there, and in fact, when d approaches infinity, it will give us alternating numbers in pascal's triangle, separated by infinite zeros.
But these are trivial.

The sum from n=1 to dk of b_n is equal to b_k.

We set d=2
the first 4 are [1,0,x,y] where x+y=-1, so there are infinitely many routes we could go. The simplest being x=-1, y=0
so, the first 6 are [1,0,-1,0,x',y'] where x'+y'=-1, using the same logic,
the first 8 are [1,0,-1,0,-1,0,x'',y''] where x''+y''=0, using x''=y''=0
... using the simplest solution
we get [1,0,-1,0,-1,0,0,0,-1,0,1,0,0,0,0,0,0,0,-1,0,1,0,1,0,0,0,0,0,0,0,...] Most likely never getting a digit above 1, due to y*=0 and b_n - b_(n+1)<=1

When we add the rule that x*=y* it might be different but I'm much too tired to pay any attention.

As a final note, I apologize if this is already a thing and I'm taking credit for it, if it is, I'd like to learn more.

Last edited by skipjack; February 12th, 2017 at 01:54 AM.
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