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A033496
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Numbers m that are the largest number in their Collatz (3x+1) trajectory.
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15
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1, 2, 4, 8, 16, 20, 24, 32, 40, 48, 52, 56, 64, 68, 72, 80, 84, 88, 96, 100, 104, 112, 116, 128, 132, 136, 144, 148, 152, 160, 168, 176, 180, 184, 192, 196, 200, 208, 212, 224, 228, 232, 240, 244, 256, 260, 264, 272, 276, 280, 288, 296, 304, 308, 312, 320, 324
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OFFSET
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1,2
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COMMENTS
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Or, possible peak values in 3x+1 trajectories: 1,2 and m=16k+4,16k+8,16k but not for all k; those 4k numbers [like m=16k+12 and others] which cannot be such peaks are listed in A087252.
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LINKS
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FORMULA
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EXAMPLE
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These peak values occur in 1, 3, 6, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 21, 22, 27, 30, 39, 44, 71, 75, 1579 [3x+1]-iteration trajectories started with different initial values. This list most probably is incomplete.
Let n be the maximum in some Collatz trajectory and let F(n), the initial fan of n, be the set of all initial values less than or equal to n whose Collatz trajectories lead to n as their maximum. Then the size of F(n) never equals 2, 4, 5, 7 or 10 (see the link).
Conjecture: Every number k > 10 occurs as the size of F(n) for some n.
Fans F(n) of size k, for all 10 < k < 355, exist for 4 <= n <= 50,000,000. The largest fan in this range, F(41163712), has size 7450.
(End)
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MATHEMATICA
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Collatz[a0_Integer, maxits_:1000] := NestWhileList[If[EvenQ[ # ], #/2, 3# + 1] &, a0, Unequal[ #, 1, -1, -10, -34] &, 1, maxits]; (* Collatz[n] function definition by Eric Weisstein *)
Select[Range[324], Max[Collatz[#]] == # &] (* T. D. Noe, Feb 28 2013 *)
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PROG
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(Haskell)
a033496 n = a033496_list !! (n-1)
a033496_list = 1 : filter f [2, 4 ..] where
f x = x == maximum (takeWhile (/= 1) $ iterate a006370 x)
(Magma) Set(Sort([Max([k eq 1 select n else IsOdd(Self(k-1)) and not IsOne(Self(k-1)) select 3*Self(k-1)+1 else Self(k-1) div 2: k in [1..5*n]]): n in [1..2^10] | Max([k eq 1 select n else IsOdd(Self(k-1)) and not IsOne(Self(k-1)) select 3*Self(k-1)+1 else Self(k-1) div 2: k in [1..5*n]]) le 2^10])) // Jaroslav Krizek, Jul 17 2016
(Python)
def a(n):
if n<2: return [1]
l=[n, ]
while True:
if n%2==0: n//=2
else: n = 3*n + 1
if n not in l:
l.append(n)
if n<2: break
else: break
return l
print([n for n in range(1, 501) if max(a(n)) == n]) # Indranil Ghosh, Apr 14 2017
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CROSSREFS
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Cf. A095384 (contains a definition of Collatz[]).
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KEYWORD
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nonn,nice,easy
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AUTHOR
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STATUS
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approved
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