A231610 The least k such that the Collatz (3x+1) iteration of k contains 2^n as the largest power of 2.

1, 2, 4, 8, 3, 32, 21, 128, 75, 512, 151, 2048, 1365, 8192, 5461, 32768, 14563, 131072, 87381, 524288, 184111, 2097152, 932067, 8388608, 5592405, 33554432, 13256071, 134217728, 26512143, 536870912, 357913941, 2147483648, 1431655765, 8589934592, 3817748707

Offset: 0

Author

T. D. Noe, Dec 02 2013

Keywords

• nonn

Comments

Very similar to A225124, where 2^n is the largest number in the Collatz iteration of A225124(n). The only difference appears to be a(8), which is 75 here and 85 in A225124. The Collatz iteration of 75 is {75, 226, 113, 340, 170, 85, 256, 128, 64, 32, 16, 8, 4, 2, 1}.

Examples

			The iteration for 21 is {21, 64, 32, 16, 8, 4, 2, 1}, which shows that 64 = 2^6 is a term. However, 32 is not the first power of two. We have to wait until the iteration for 32, which is {32, 16, 8, 4, 2, 1}, to see 32 = 2^5 as the first power of two.
		

Crossrefs

Cf. A010120, A054646 (similar sequences).

Cf. A135282, A232503 (largest power of 2 in the Collatz iteration of n).

Cf. A225124.

Programs

Mathematica

Collatz[n_?OddQ] := 3*n + 1; Collatz[n_?EvenQ] := n/2; nn = 21; t = Table[-1, {nn}]; n = 0; cnt = 0; While[cnt < nn, n++; q = Log[2, NestWhile[Collatz, n, Not[IntegerQ[Log[2, #]]] &]]; If[q < nn && t[[q + 1]] == -1, t[[q + 1]] = n; cnt++]]; t

Formula

a(n) = 2^n for odd n.