If k is an odd integer, then 3k + 1 is even, so we can write 3k + 1 = 2ak′, with k' odd and a ≥ 1. We define a function f from the set I of odd integers into itself, called the Syracuse Function, by taking f (k) = k′ (sequence A075677 in OEIS).
Some properties of the Syracuse function are:
f (4k + 1) = f (k) for all k in I.
For all p ≥ 2 and h odd, f p - 1(2 p h - 1) = 2 3 p - 1h - 1 (see here for the notation).
For all odd h, f (2h - 1) ≤ (3h - 1)/2
The Syracuse Conjecture is that for all k in I, there exists an integer n ≥ 1 such that f n(k) = 1. Equivalently, let E be the set of odd integers k for which there exists an integer n ≥ 1 such that f n(k) = 1. The problem is to show that E = I. The following is the beginning of an attempt at a proof by induction:
1, 3, 5, 7, and 9 are known to exist in E. Let k be an odd integer greater than 9. Suppose that the odd numbers up to and including k - 2 are in E and let us try to prove that k is in E. As k is odd, k + 1 is even, so we can write k + 1 = 2ph for p ≥ 1, h odd, and k = 2ph-1. Now we have:
If p = 1, then k = 2h - 1. It is easy to check that f (k) < k , so f (k) ∈ E; hence k ∈ E.
If p ≥ 2 and h is a multiple of 3, we can write h = 3h′. Let k′ = 2p + 1h′ - 1; we have f (k′) = k , and as k′ < k , k′ is in E; therefore k = f (k′) ∈ E.
If p ≥ 2 and h is not a multiple of 3 but h ≡ (-1)p mod 4, we can still show that k ∈ E. (Cf.)
The problematic case is that where p ≥ 2 , h not multiple of 3 and h ≡ (-1)p+1 mod 4. Here, if we manage to show that for every odd integer k′, 1 ≤ k′ ≤ k-2 ; 3k′ ∈ E we are done. (Cf.).
and btw i want a snowboard and some sand dunes
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I have a math teacher at my school that can do anything math related and make it look like using an easy bake....I'm taking that crap and showing him, and if he does something with it I am going to slap the taste out of his mouth. Who has time for stuff like that? Any way those dunes look pimpin, I want to take my 525 and bury the sob there. Since the wreck its been crap, thats for another time though lol.