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%I #69 Jul 12 2021 03:10:37
%S 0,0,1,0,1,1,0,1,1,3,0,1,1,1,1,0,1,1,3,1,5,0,1,1,3,1,5,3,0,1,1,3,1,5,
%T 3,7,0,1,1,3,1,5,3,7,1,0,1,1,3,1,1,3,7,1,9,0,1,1,3,1,5,3,7,1,3,5,0,1,
%U 1,3,1,5,3,7,1,9,5,5,0,1,1,3,1,5,3,7,1,9,5,1
%N Square array T(m,n), read by ascending antidiagonals. Let f(k) = k/2 if k is even, otherwise ((2*n+1)*k+2*r+1)/2, r is the smallest integer greater than -1, where m = f^j(m) for j > 0 exists and is determined in A345228, T(m,n) is the smallest number reached in the cyclic trajectory of m = f^j(m). f^j(m) means j times recursion into f(m).
%C This sequence, together with A345228, provides information regarding generalized Collatz functions. (Replace 3*k+1 in the standard Collatz function with a more general a*k+b; then a = 1+2*n and b = 1+2*A345228(m,n).) A345228 tells us which m are part of a cyclic orbit but not if these are part of the same cycle. This sequence identifies each distinct cycle with a different number. Example: If A345228(m1,n) = A345228(m2,n) we know m1 and m2 are part of a cycle but not necessarily the same cycle. If T(m1,n) <> T(m2,n) we know m1 and m2 are not in the same cycle.
%C The value of n appears to have only a small effect in this sequence and in a majority of cases we find T(m,n) = A000265(m) holds true. This is surprising, given how n is involved in the definition.
%H Thomas Scheuerle, <a href="/A343858/a343858.svg">Yellow dot if T(m,n) <> A000265(n); m = 1..100; n = 1..20</a>.
%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>.
%F T((1+2*n)*m,n)/T(m,n) = 1+2*n.
%F T((1+2*(n-b))*m,n)/T(m,n) = 1+2*(n-b). 0 <= b <= n. This formula is only for the majority of cases true if b > 0. For each column m are some rows n where an exception will be seen.
%F T(m,n) <= A000265(m) (largest odd divisor of m).
%F T(m,n) = A000265(m) For the majority of all n.
%e Twelve initial terms of rows 0-10 are listed below:
%e n |m->
%e 0: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 1, 11, ...
%e 1: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 5, ...
%e 2: 0, 1, 1, 1, 1, 5, 3, 7, 1, 3, 5, 1, ...
%e 3: 0, 1, 1, 3, 1 5, 3, 7, 1, 9, 5, 11, ...
%e 4: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, ...
%e 5: 0, 1, 1, 3, 1, 1, 3, 7, 1, 9, 1, 11, ...
%e 6: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 1, ...
%e 7: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, ...
%e 8: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, ...
%e 9: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, ...
%e 10: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, ...
%e Example: T(3,4) = 3 -> f(n): k/2; (9*k+21)/2. This is because r = A345228(3,4) = 10 and 2*10+1 = 21.
%e f(3) = 24, f(24) = 12, f(12) = 6, f(6) = 3, f(3) = 24, ....
%e The smallest number in this cycle is 3.
%Y Cf. A344583, A345228, A000265.
%K nonn,tabl
%O 0,10
%A _Thomas Scheuerle_, Jun 14 2021
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