A372453 a(n) = A372443(n) - A086893(1+A372447(n)).

6, -12, 10, -6, -14, 22, -52, 36, 6, -76, 18, -58, 20, -38, -78, 54, -260, 104, -46, 38, 36, -58, 84, -22, 138, -134, -286, 254, -984, 58, 2, -1362, -336, -276, 92, -16, 8, 2, -18, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Author

Antti Karttunen, May 05 2024

Keywords

• sign

Comments

These are the differences obtained when the term of A086893 that has the same binary length as A372443(n) is subtracted from the latter. Here A372443(n) gives the n-th iterate of 27 with Reduced Collatz-function R, where R(n) = A000265(3*n+1).

Note that for all n >= 1, R(A086893(2n-1)) = 1, and R(A086893(2n)) = 5 (with R(5) = 1), so the first zero here, a(39) = 0 indicates that the iteration will soon have reached the terminal 1, and indeed, A372443(41) = 1.

Examples

			The term of A086893 that has same binary length as A372443(0) = 27 is 21 [as 21 = 10101_2 in binary, and 27 = 11011_2 in binary], therefore a(0) = 27-21 = 6.
The term of A086893 that has same binary length as A372443(1) = 41 is 53, therefore a(1) = 41-53 = -12.
		

Links

• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A000265, A000523, A086893, A372443.

Cf. also A372446.

Programs

PARI

A000265(n) = (n>>valuation(n,2));
A000523(n) = logint(n,2);
A086893(n) = (if(n%2, 2^(n+1), 2^(n+1)+2^(n-1))\3);
A372443(n) = { my(x=27); while(n, x=A000265(3*x+1); n--); (x); };
A372453(n) = { my(x=A372443(n)); (x - A086893(1+A000523(x))); };

Formula

a(n) = A372443(n) - A086893(1+A000523(A372443(n))).