A345228 -id:A345228 - cp's OEIS frontend

A343858 Square array T(m,n), read by ascending antidiagonals. Let f(k) = k/2 if k is even, otherwise ((2n+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).

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, 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, 1, 3, 1, 5, 3, 7, 1, 9, 5, 5, 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 1

Offset: 0

Thomas Scheuerle, Jun 14 2021

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.

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.

			Twelve initial terms of rows 0-10 are listed below:
   n |m->
   0: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 1, 11, ...
   1: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5,  5, ...
   2: 0, 1, 1, 1, 1, 5, 3, 7, 1, 3, 5,  1, ...
   3: 0, 1, 1, 3, 1  5, 3, 7, 1, 9, 5, 11, ...
   4: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, ...
   5: 0, 1, 1, 3, 1, 1, 3, 7, 1, 9, 1, 11, ...
   6: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5,  1, ...
   7: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, ...
   8: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, ...
   9: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, ...
  10: 0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, ...
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.
f(3) = 24, f(24) = 12, f(12) = 6, f(6) = 3, f(3) = 24, ....
The smallest number in this cycle is 3.

Thomas Scheuerle, Yellow dot if T(m,n) <> A000265(n); m = 1..100; n = 1..20.
Index entries for sequences related to 3x+1 (or Collatz) problem.

Cf. A344583, A345228, A000265.

T((1+2*n)*m,n)/T(m,n) = 1+2*n.
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.
T(m,n) <= A000265(m) (largest odd divisor of m).
T(m,n) = A000265(m) For the majority of all n.

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