A277962 Least k such that A006577(k) = A006577(n) + A006577(n+1).

2, 6, 12, 3, 34, 49, 9, 72, 98, 18, 25, 28, 33, 39, 36, 7, 57, 406, 65, 11, 72, 86, 98, 114, 114, 129, 913, 153, 153, 171, 27, 172, 203, 33, 39, 270, 270, 295, 270, 290, 290, 305, 361, 57, 57, 386, 73, 78, 481, 481, 78, 72, 514, 20174, 609, 641, 641, 641, 641

Offset: 1

Author

Michel Lagneau, Nov 06 2016

Keywords

• nonn

Comments

A006577(n) is the number of halving and tripling steps to reach 1 in the '3x+1' problem.

The distinct squares in the sequence are 9, 25, 36, 49, 169, 361, ...

The distinct primes in the sequence are 2, 3, 7, 11, 31, 41, 47, 71, 73, 97, 103, ...

Examples

			a(5)=34 because A006577(34) = 13 = A006577(5) + A006577(6) = 5 + 8.
		

Links

• Michel Lagneau, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006577.

Programs

Maple

nn:=3*10^6:U:=array(1..nn):V:=array(1..nn):
for i from 1 to nn do:
m:=i:it0:=0:
   for j from 1 to nn while(m<>1) do:
    if irem(m, 2)=0
     then
     m:=m/2:it0:=it0+1:
     else
     m:=3*m+1:it0:=it0+1:
    fi:
   od:
   U[i]:=it0:
  od:
   for n from 1 to 60 do:
   ii:=0:
    for k from 1 to nn while(ii=0) do:
     if U[k]=U[n]+ U[n+1]
      then
      ii:=1:printf(`%d, `, k):
      else
     fi:
    od:
od:

Mathematica

f:=Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#!=1&]]-1,{n,3*10^6}];Do[k=1;While[f[[k]]!=f[[m]]+f[[m+1]],k++];Print[m," ",k],{m,1,60}]