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Search: id:A119733
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| A119733 |
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Offsets of the terms of the nodes of the reverse Collatz function. |
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+0
1
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| 0, 1, 2, 5, 4, 7, 10, 19, 8, 11, 14, 23, 20, 29, 38, 65, 16, 19, 22, 31, 28, 37, 46, 73, 40, 49, 58, 85, 76, 103, 130, 211, 32, 35, 38, 47, 44, 53, 62, 89, 56, 65, 74, 101, 92, 119, 146, 227, 80, 89, 98, 125, 116, 143, 170, 251, 152, 179, 206, 287, 260, 341, 422, 665, 64, 67
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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Create a binary tree starting with x. To follow 0 from the root, apply f(x)=2x. To follow 1, apply g(x)=(2x-1)/3. For example, starting with x, the string 010 {also known as f(g(f(x))) }, you would get (8x-2)/3. These expressions represent the reverse Collatz function and will provide numbers who's Collatz path may include x. These expressions will all be of the form (2^a*x-b)/3^c. This sequence concerns b. What makes b interesting is that if you draw the tree, each level of the tree will have the same sequence of values for b. The root of the tree x, can be written as (2^0*x-0)/3^0, which has the first value for b. Each subsequent level contains twice as many values of b.
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FORMULA
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a(0)=0, a(2i)=2*a(i)+3^[the number 1's in the binary representation of i], a(2i-1)=2*a(i)
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EXAMPLE
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a(2)=1=2*0+3^0 since 0 written in binary contains no 1's.
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PROGRAM
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# perl, sorry :-( # call with n to get 2^n values $depth=shift; sub funct { my ($i, $b, $c) = @_; if ($i < $depth) { funct($i+1, $b*2, $c); funct($i+1, 2*$b+$c, $c*3); } else { print "$b, "; } } funct(0, 0, 1); print " ";
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CROSSREFS
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Adjacent sequences: A119730 A119731 A119732 this_sequence A119734 A119735 A119736
Sequence in context: A154811 A036237 A015948 this_sequence A111570 A057954 A155896
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KEYWORD
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nonn,uned
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AUTHOR
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William Entriken (att(AT)phor.net), Jun 14 2006
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