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A082983
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Odd numbers which lead to 1 in the 3x+1 problem, generated by a particular "least-first" greedy algorithm (see program code).
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0
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1, 5, 3, 13, 17, 11, 7, 9, 21, 29, 19, 25, 33, 37, 45, 49, 53, 35, 23, 15, 61, 65, 43, 57, 69, 77, 51, 81, 85, 93, 101, 67, 89, 59, 39, 113, 75, 117, 133, 141, 149, 99, 157, 173, 115, 153, 177, 181, 197, 131, 87, 205, 209, 139, 185, 123, 213, 229, 237, 241, 245, 163, 217
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OFFSET
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0,2
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COMMENTS
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It is suspected but not proved that all odd integers are in the sequence - this is equivalent to whether all numbers reach 1 in the 3x+1 problem. The program code given below does not actually represent infinite sets, but the result is the same since the smallest remaining member of each sibling-set is always present.
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LINKS
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EXAMPLE
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The second term is 5 because if we take m = 1 (our starting point), generate all numbers m * 2^k, k >= 1, subtract 1 and divide by 3 for those which give an integer result, we get the set {5,21,85, ...} (sequence A002450), which we call the "children" of 1, and 5 is the smallest member of that set. Next we replace 5 in the set by 5's children {3,13,53,213, ...} (sequence A072197), forming a new working set of generated numbers that are not yet in the sequence. The next term is 3 because the smallest member is 3. 3 has no children, so we simply remove 3 from the working set and the next term is 13. [Edited by Peter Munn, Jun 20 2021]
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PROG
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#!/usr/bin/perl @list = ( 1 ); while (1) { $n = shift @list; print "$n "; # next sibling push(@list, 4*$n + 1); # first child if (($n % 3) == 1) { $n = ($n*4 - 1)/3; while ($n && (($n % 2) == 0)) { $n /= 2; } push(@list, $n) unless ($n <= 1); }
elsif (($n % 3) == 2) { $n = ($n*2 - 1)/3; while ($n && (($n % 2) == 0)) { $n /= 2; } push(@list, $n) unless ($n <= 1); } #else do nothing, since == 0 mod 3 has no children # Inefficient - should have heap insertion sort. @list = sort numeric @list; } sub numeric { $a <=> $b; }
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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