Fibonacci but with Subtraction still stays Positive (Posted on 2023-02-15)

	Submitted by Brian Smith
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Let D[2]=x then the sequence continues as D[3] = 1 - x D[4] = -1 + 2x D[5] = 2 - 3x D[6] = -3 + 5x D[7] = 5 - 8x D[8] = -8 + 13x D[9] = 13 - 21x Its pretty easy to see the pattern with Fibonacci numbers. So let them be represented by F[n] with F[0]=0 and F[1]=1. Then the sequence looks like this: D[3] = F[1] - F[2]x D[4] = -F[2] + F[3]x D[5] = F[3] - F[4]x D[6] = -F[4] + F[5]x D[7] = F[5] - F[6]x D[8] = -F[6] + F[7]x D[9] = F[7] - F[8]x Now it is more clear we have two rules, one for odd terms and one for even terms: D[2m] = -F[2m-2] + F[2m-1]x D[2m+1] = F[2m-1] - F[2m]x Both of these must be greater than zero. So solving each inequality for x we can create the compound inequality F[2m-2]/F[2m-1] < x < F[2m-1]/F[2m] The ratio of consecutive terms of the Fibonacci sequence, F[2m+1]/F[2m], approach the golden ratio, phi = (1+sqrt(5))/2. So if we apply that limit to the ends of our inequality, then we get that potential x can only be the single value 1/phi = (-1+sqrt(5))/2. 1/phi obeys the relation 1 - 1/phi = (1/phi)^2. So with that we can write the sequence with x = 1/phi as: D[3] = (1/phi)^2 D[4] = (1/phi)^3 D[5] = (1/phi)^4 D[6] = (1/phi)^5 D[7] = (1/phi)^6 D[8] = (1/phi)^7 D[9] = (1/phi)^8 Clearly this is a geometric series, so we know the terms are all positive. This establishes D[2] = (-1+sqrt(5))/2 as the single value of D[2] for which the sequence D[n] is always positive.

	Subject	Author	Date
	Puzzle Thoughts	K Sengupta	2023-02-22 02:25:00
	solution	Charlie	2023-02-15 17:13:49
	Solution	Larry	2023-02-15 13:43:48