A075477 Number of iteration that first becomes smaller than the initial value if Collatz-function (A006370) is iterated, starting with numbers of form 64n+15. Corresponds to selection of every 16th term from A074474.

12, 14, 12, 22, 12, 14, 12, 20, 12, 14, 12, 22, 12, 14, 12, 17, 12, 14, 12, 20, 12, 14, 12, 40, 12, 14, 12, 58, 12, 14, 12, 17, 12, 14, 12, 33, 12, 14, 12, 33, 12, 14, 12, 25, 12, 14, 12, 17, 12, 14, 12, 33, 12, 14, 12, 27, 12, 14, 12, 40, 12, 14, 12, 17, 12, 14, 12, 69, 12

Offset: 0

Author

Labos Elemer, Sep 23 2002

Keywords

• nonn

Comments

Remark that initial values of form 64m+r, if r={3,11,19,27,35,43,51,55} provide first-sink-lengths {7,9,7,9,7,9,7,9} respectively; e.g. {64k+19,192k+58,96k+29,288k+88,144k+44,72k+22,36k+11} submerge first below initial value at the 7th term,36k+11<64k+19.

Examples

			n=0: 64n+15=15,list={15,46,23,70,35,106,53,160,80,40,20,10..}, i.e. the 12th term is the first that <15, the initial value.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..16384
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A074473, A074474, A075476-A075482, A006370.

Programs

PARI

A006370(n) = if(n%2, 3*n+1, n/2);
A074473(n) = if(1==n,n,my(org_n=n); for(i=1,oo,if(nA006370(n)));
A075477(n) = A074473((64*n)+15); \\ Antti Karttunen, Oct 09 2018

Formula

a(n) = A074473(64n+15), n=0..256. [corrected by Antti Karttunen, Oct 09 2018]