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A284668
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Numbers that have the largest Collatz total stopping time of all numbers below 10^n. The smallest number is chosen in case of ties.
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0
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9, 97, 871, 6171, 77031, 837799, 8400511, 63728127, 670617279, 9780657630, 75128138247, 989345275647, 7887663552367, 80867137596217, 942488749153153, 7579309213675935, 93571393692802302, 931386509544713451
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OFFSET
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1,1
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COMMENTS
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Collatz stopping time is defined as the number of steps that a number n takes to converge to 1 using one of the following steps:
0) if n is 1, stop.
1) if n is even, divide n by 2 (n/2).
2) if n is odd, multiply n by 3 and add 1 (3n+1).
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LINKS
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Gary T. Leavens and Mike Vermeulen, 3x+1 Search Programs, Computers & Mathematics with Applications. 24 (11): 79-99 (1992).
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FORMULA
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a(n) = max{i} (steps(i) for i in range from 1 to 10^n-1).
max(i) returns the i with the maximum steps(i) value.
where steps(n) is defined as follows
steps(n)= 0 if n=1.
1+steps(n/2) if n is even.
1+steps(3*n+1) if n is odd.
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EXAMPLE
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For n=1, steps(1) to steps(9) take the following values: 0, 1, 7, 2, 5, 8, 16, 3, 19; the maximum of all those is 19 which occurs for steps(9) therefore a(1)=9.
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MATHEMATICA
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Table[Last@Ordering@Array[If[#>1, #0@If[OddQ@#, 3#+1, #/2]+1, 0]&, 10^k], {k, 4}] (* Giorgos Kalogeropoulos, Apr 01 2021 *)
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PROG
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(Python 3.4)
def steps(n):
if n==1:
return 0
else:
if (n%2)==0:
return 1+steps(int(n/2))
else:
return 1+steps(3*n+1)
def max_steps(i):
a=max([[i, steps(i)] for i in range(1, 10**(i))], key=lambda x:x[1])
return a[0]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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