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Irregular rows of odd numbers that produce n even numbers in their Collatz iteration.
+10
6
5, 3, 21, 13, 85, 17, 53, 11, 35, 113, 341, 7, 23, 69, 75, 213, 227, 15, 45, 141, 151, 453, 1365, 9, 29, 93, 277, 301, 853, 909, 19, 61, 181, 201, 565, 605, 1813, 5461, 37, 117, 369, 373, 401, 403, 1109, 1137, 1205, 3413, 3637, 25, 77, 81, 241, 245, 267, 725
COMMENTS
It is conjectured that every odd number greater than 1 eventually appears in this sequence. The smallest and largest terms in row n are A199637(n) and A199638(n). The number of terms in row n is A131450(n) for n > 3.
MATHEMATICA
Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 16; t = Table[{}, {nn}]; Do[len = Length[Select[Collatz[n], EvenQ]]; If[0 < len <= nn, AppendTo[t[[len]], n]], {n, 1, 25000, 2}]; t
The number of halving steps of the Collatz 3x+1 map to reach 1 starting from 2n-1.
+10
4
0, 5, 4, 11, 13, 10, 7, 12, 9, 14, 6, 11, 16, 70, 13, 67, 18, 10, 15, 23, 69, 20, 12, 66, 17, 17, 9, 71, 22, 22, 14, 68, 19, 19, 11, 65, 73, 11, 16, 24, 16, 70, 8, 21, 21, 59, 13, 67, 75, 18, 18, 56, 26, 64, 72, 45, 10, 23, 15, 23, 61, 31, 69, 31, 77, 20, 20, 28, 58, 28, 12, 66, 74, 74, 17
MAPLE
A006370 := proc(n) if type(n, 'even') then n/2; else 3*n+1 ; end if; end proc:
A006577 := proc(n) a := 0 ; x := n ; while x > 1 do x := A006370(x) ; a := a+1 ; end do; a ; end proc:
A006667 := proc(n) a := 0 ; x := n ; while x > 1 do if type(x, 'even') then x := x/2 ; else x := 3*x+1 ; a := a+1 ; end if; end do; a ; end proc:
# second Maple program:
b:= proc(n) option remember; `if`(n=1, 0,
1+b(`if`(n::even, n/2, (3*n+1)/2)))
end:
a:= n-> b(2*n-1):
MATHEMATICA
b[n_] := b[n] = If[n == 1, 0, 1 + b[If[EvenQ[n], n/2, (3n+1)/2]]];
a[n_] := b[2n-1];
Consider the version of the Collatz or 3x+1 problem where x -> x/2 if x is even, x -> (3x+1)/2 if x is odd. Define the stopping time of x to be the number of steps needed to reach 1. Sequence gives the number of integers x with stopping time n.
+10
3
1, 1, 1, 1, 2, 3, 4, 5, 6, 8, 12, 18, 24, 31, 39, 50, 68, 91, 120, 159, 211, 282, 381, 505, 665, 885, 1187, 1590, 2122, 2829, 3765, 5014, 6682, 8902, 11878, 15844, 21122, 28150, 37536, 50067, 66763, 89009, 118631, 158171, 210939, 281334, 375129
COMMENTS
The Mathematica function StoppingTime[n] is the length of the Collatz sequence starting at n before reaching 1.
Suppose we have a list L of the numbers with StoppingTime n. Then the list LL of StoppingTime n+1 can be produced as: First. Add to LL all numbers in L multiplied by 2. Second. For the numbers x now in LL, if x == 1 (mod 3), AppendTo LL the number (x-1)/3 (if (x-1)/3 != 1). These two steps make LL complete.
I think the offset, examples, formula and code are all off by 1 -- they all treat the stopping time of 1 to be 1, rather than 0. - David Applegate, Oct 16 2008 a(n+1), n >= 0, is the row length of A248573(n,m) (Collatz-Terras tree). For the first differences see A131450(n+1), but with A131450(2) = 1 (the number of 2 (mod 3) numbers in row n, for n >= 0, of A248573). - Wolfdieter Lang, May 04 2015
FORMULA
Conjecture: lim_{n->oo} a(n) = a(n-1)*4/3. - Joe Slater, Jan 27 2024
EXAMPLE
StoppingTime = 1: L = {1}, a(1)=1.
StoppingTime = 2: L = {2}, a(2)=1.
StoppingTime = 3: L = {4}, a(3)=1.
StoppingTime = 4: L = {8}, a(4)=1.
StoppingTime = 5: L = {5, 16}, a(5)=2. First, LL = {10, 32} (= 2*L). Second, 10 == 1 (mod 3), so we AppendTo LL also (10-1)/3 = 3. We get LL = {3, 10, 32}. So a(6) = 3.
MATHEMATICA
(*** Program #1 ***) For[v = 1, v <= 12, v++, lst = {}; For[n = 1, n < 2^v, n++, If[StoppingTime[n] == v, AppendTo[lst, n]]]; Print[lst]; Print[Length[lst]]; ]
(*** Program #2 ***) lst1 = {1}; For[v = 1, v <= 12, v++, L1 = Length[lst1]; Print["Number of numbers with StoppingTime ", v, ": ", L1]; Print["List of numbers: ", lst1]; (* Numbers with StoppingTime n *) Print["Control of StoppingTime: ", Map[StoppingTime, lst1]]; (* Controll *) Print[""]; lst2 = 2 lst1; For[i = 1, i <= L1, i++, x = (lst2[[i]] - 1)/3; If[IntegerQ[x] && x != 1, AppendTo[lst2, x]]; ]; lst1 = Sort[lst2]; ]
(*** Program #3 ***) lst0 = {}; lst1 = {1}; For[v = 1, v <= 35, v++, L1 = Length[lst1]; AppendTo[lst0, L1]; lst2 = 2 lst1; For[i = 1, i <= L1, i++, x = (lst2[[i]] - 1)/3; If[IntegerQ[x], AppendTo[lst2, x]]; ]; lst1 = Complement[lst2, {1}]; ]; lst0
PROG
(Perl) # code to calculate terms after a(4):
@x=(8, 0); for($n=5; $n<=60; $n++){do{$q=2*shift(@x); push(@x, ($q-1)/3)if($q%3==1); push @x, $q}while $q; print($#x, ", "); } # Carl R. White, Oct 03 2006 (PARI) first(N) = my(a=Vec([1, 1, 1, 1, 2], N), p=[], q=[5]); for(n=6, N, my(r=List()); foreach(p, x, listput(r, 4*x+1); if(1==x%6, listput(r, x+(x-1)/3))); foreach(q, x, if(5==x%6, listput(r, x-(x+1)/3))); a[n]=a[n-1]+#r; p=q; q=Vec(r)); a; \\ Ruud H.G. van Tol, Aug 14 2024
AUTHOR
Bo T. Ahlander (ahlboa(AT)isk.kth.se), Mar 29 2001
Odd numbers producing 10 even numbers in the Collatz iteration.
+10
2
MATHEMATICA
Collatz[n_]:=NestWhileList[If[EvenQ[#], #/2, 3 #+1]&, n, #>1&]; t={}; Do[If[Length[Select[Collatz[n], EvenQ]] == 10, AppendTo[t, n]], {n, 1, 100000, 2}]; t
Odd numbers producing 20 even numbers in the Collatz iteration.
+10
2
43, 131, 133, 397, 405, 433, 435, 441, 475, 1237, 1251, 1285, 1301, 1313, 1325, 1339, 1425, 1427, 1431, 1433, 3861, 3925, 3939, 3941, 3981, 4017, 4019, 4043, 4277, 4293, 4297, 4301, 11605, 11829, 12053, 12131, 12133, 12853, 12885, 12893, 12913, 12931, 36181
COMMENTS
For n <10000000, more terms: 36405, 38677, 38741, 38797, 38833, 38835, 116053, 116501, 349525.
MATHEMATICA
Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; t = {}; Do[If[Length[Select[Collatz[n], EvenQ]] == 20, AppendTo[t, n]], {n, 1, 100000, 2}]; t
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