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A288251
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k-th term of n in the 3x+1 problem, k >= 1. Square array A(n, k) read by antidiagonals downwards.
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0
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4, 2, 1, 1, 4, 10, 4, 2, 5, 2, 2, 1, 16, 1, 16, 1, 4, 8, 4, 8, 3, 4, 2, 4, 2, 4, 10, 22, 2, 1, 2, 1, 2, 5, 11, 4, 1, 4, 1, 4, 1, 16, 34, 2, 28, 4, 2, 4, 2, 4, 8, 17, 1, 14, 5, 2, 1, 2, 1, 2, 4, 52, 4, 7, 16, 34, 1, 4, 1, 4, 1, 2, 26, 2, 22, 8, 17, 6, 4, 2, 4
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OFFSET
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1,1
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LINKS
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FORMULA
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T(m, n) = A(n, m-n+1), 1 <= n <= m, with A(n, k) = C^{[k]}(n), n >= 1, k >= 1, with iterations of the Collatz map C: n -> 3*n+1 if n is odd and n -> n/2 if n is even. - Wolfdieter Lang, Jul 13 2017
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EXAMPLE
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The array A(n, k) starts:
n\k | 1 2 3 4 5 6 7 8 9 10 11 12 ...
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1 | 4 2 1 4 2 1 4 2 1 4 2 1
2 | 1 4 2 1 4 2 1 4 2 1 4 2
3 | 10 5 16 8 4 2 1 4 2 1 4 2
4 | 2 1 4 2 1 4 2 1 4 2 1 4
5 | 16 8 4 2 1 4 2 1 4 2 1 4
6 | 3 10 5 16 8 4 2 1 4 2 1 4
7 | 22 11 34 17 52 26 13 40 20 10 5 16
8 | 4 2 1 4 2 1 4 2 1 4 2 1
9 | 28 14 7 22 11 34 17 52 26 13 40 20
10 | 5 16 8 4 2 1 4 2 1 4 2 1
...
The triangle T(m, n) starts:
m\n | 1 2 3 4 5 6 7 8 9 10 ...
------------------------------------
1 | 4
2 | 2 1
3 | 1 4 10
4 | 4 2 5 2
5 | 2 1 16 1 16
6 | 1 4 8 4 8 3
7 | 4 2 4 2 4 10 22
8 | 2 1 2 1 2 5 11 4
9 | 1 4 1 4 1 16 34 2 28
10 | 4 2 4 2 4 8 17 1 14 5
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PROG
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(PARI) trajectory(n, terms) = my(i=0, x=n); while(1, if(x%2==0, x=x/2, x=3*x+1); print1(x, ", "); i++; if(i==terms, break))
array(n, k) = for(x=1, n, trajectory(x, k); print(""))
array(10, 12) \\ print initial 10 rows and 12 columns of array
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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