# Possible improvements to this Syracuse (3x+1)/2 graph?

This algorithm produces the Syracuse disjoint tree graph without any duplicates. No need for Union, For, and While. The function α is based on this OEIS sequence. The function β is a wrapper for IntegerExponent. Related math.SE question.

    α[n_] := 3 n - (5 + (-1)^n)/2
β[m_] := IntegerExponent[m, 2]
a = Table[Join[
{Table[x -> (x = (3 x + 1)/2), {β[(x = α[j]) + 1] - 1}]},
{x -> (3 x + 1)/2^β[3 x + 1]}
],
{j, 1, 150}];
Graph[Flatten[a]]


Fifteen sequences: , 150 sequences: Edit It seems I was abusing the set-builder notation, so my question at math.SE will not parallel the Mathematica statements. So, this question remains: Is there any way to improve the Table expression? It was suggested that NestList[] might be the ticket.

• NestList[] might make for a cleaner implementation here, I think. – J. M.'s technical difficulties May 31 '15 at 2:38

You're setting x as a side-effect and that (I believe) makes your code difficult to follow. This one is equivalent using a "more functional" programing style.

As @Guesswhoitis suggested, NestList[] is your friend.

a[n_] := 3 n - (5 + (-1)^n)/2
b[m_] := IntegerExponent[m, 2]
nextSeq[n_] := (#/2^b@#) &[1 + 3 n]
full[j_] := NestList[nextSeq, a@j, b[a@j + 1]]
Graph[DirectedEdge @@@ Flatten[Partition[#, 2, 1] & /@ full /@ Range@15, 1]] I don't know anything about set-builder notation, but perhaps the following is an approximation:

$$\{(f^k(a(j)), f^{k+1}(a(j))) \ | \ \{ k,j\} \in \mathbb{Z}\ \wedge\ \ 0\le\ k \le\ b(a(j)+1) - 1\ \wedge 0\le\ j \le\ n \}$$

($f$ is the nextSeq[ ] function in the above snippet)

GraphicsGrid@
Partition[
Graph[DirectedEdge @@@Flatten[Partition[#, 2, 1] &/@ full/@ Range@#,1]]&/@ Range@50,
10] • "setting x as a side-effect" - after all, that is how one would do Collatz in a procedural language. Thanks for showing OP the NestList[] route! (Tho, what I had in mind involved Partition[] as well as NestList[].) – J. M.'s technical difficulties Jun 1 '15 at 23:52
• Still computer-less, unfortunately, but let me run through my scratch paper again, and I'll get back to you with a sketch. – J. M.'s technical difficulties Jun 2 '15 at 0:59
• @FredKline I posted a far better version that doesn't need a separated logic for the last element – Dr. belisarius Jun 2 '15 at 3:07
• @Guesswhoitis. there you've it :) – Dr. belisarius Jun 2 '15 at 5:01
• Ah, that's more or less what I had in my paper. Thanks for following through, and I'm sorry I can't upvote again. – J. M.'s technical difficulties Jun 2 '15 at 5:16