A223699 Least number whose Collatz (3x+1) iteration has its maximum value at position n (counting from 1).

1, 2, 4, 8, 3, 32, 21, 20, 13, 80, 15, 7, 96, 69, 68, 45, 93, 19, 61, 56, 37, 51, 72, 49, 39, 33, 43, 133, 79, 260, 115, 349, 255, 127, 27, 157, 135, 279, 123, 421, 375, 727, 219, 723, 447, 295, 740, 493, 439, 591, 657, 1281, 1159, 877, 759, 615, 519, 1603

Offset: 1

Author

T. D. Noe, Mar 26 2013

Keywords

• nonn

Comments

The maximum values are in A223700.

Examples

			The Collatz iteration of 15 is {15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1}. The maximum is 160, which occurs at position 11, counting from the right. Hence, a(11) = 15 because no number smaller than 15 has its maximum value at the 11 position.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..800

Crossrefs

Cf. A070165 (Collatz iteration of n), A223700.

Programs

Mathematica

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; t= Table[c = Reverse[Collatz[n]]; Position[c, Max[c]][[1, 1]], {n, 10000}]; t2 = {}; n = 0; While[n++; p = Position[t, n, 1, 1]; p != {}, AppendTo[t2, p[[1,1]]]]; t2