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0, 2, 0, 5, 3, 8, 2, 11, 6, 14, 0, 17, 9, 20, 5, 23, 12, 26, 3, 29, 15, 32, 8, 35, 18, 38, 2, 41, 21, 44, 11, 47, 24, 50, 6, 53, 27, 56, 14, 59, 30, 62, 0, 65, 33, 68, 17, 71, 36, 74, 9, 77, 39, 80, 20, 83, 42, 86, 5, 89, 45, 92, 23, 95, 48, 98, 12, 101, 51, 104, 26, 107, 54, 110, 3
(list;
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internal format)
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OFFSET
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0,2
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COMMENTS
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All positive integers eventually reach 1 in the Collatz problem iff all nonnegative integers eventually reach 0 with repeated application of this map, i.e., if for all n, the sequence n, a(n), a(a(n)), a(a(a(n))), ... eventually hits 0.
0 <= a(n) <= (3n+1)/2, with the upper bound being achieved for all odd n.
The positions of the zeros are given by A020988 = (2/3)*(4^n-1). This is because if n = (2/3)*(4^k-1), then m = 2n+1 = (1/3)*(4^(k+1)-1), and 3m+1 = 4^(k+1) is a power of 4. - Howard A. Landman, Mar 14 2010
Subsequence of A025480, a(n) = A025480(3n+1), i.e., A025480 = 0,[0],1,0,[2],1,3,[0],4,2,[5],1,6,[3],7,0,[8],4,9,[2],10,5,[11],1,12,[6],13,3,[14],... with elements of A173732 in brackets. - Paul Tarau, Mar 21 2010
Original name: "A compression of the Collatz (or 3x+1) sequence considered as a map from odd numbers to odd numbers." - Michael De Vlieger, Oct 07 2019
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LINKS
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EXAMPLE
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a(0) = 0 because 2n+1 = 1 (the first odd number), 3*1 + 1 = 4, dividing all powers of 2 out of 4 leaves 1, and (1-1)/2 = 0.
a(1) = 2 because 2n+1 = 3, 3*3 + 1 = 10, dividing all powers of 2 out of 10 leaves 5, and (5-1)/2 = 2.
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MATHEMATICA
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Array[(#/2^IntegerExponent[#, 2] - 1)/2 &[6 # + 4] &, 75, 0] (* Michael De Vlieger, Oct 06 2019 *)
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PROG
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(C) #include <stdio.h> main() { int k, m, n; for (k = 0; ; k++) { n = 2*k + 1 ; m = 3*n + 1 ; while (!(m & 1)) { m >>= 1 ; } printf("%d, ", ((m - 1) >> 1)); } }
(Haskell)
a173732 n = a173732_list !! n
a173732_list = f $ tail a025480_list where f (x : _ : _ : xs) = x : f xs
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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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EXTENSIONS
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STATUS
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approved
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