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

1,1

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, ...

Links

Michel Lagneau, Table of n, a(n) for n = 1..10000

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

Example

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

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}]

Crossrefs

Cf. A006577.

Sequence in context: A188894 A355416 A054579 * A209386 A205103 A367575

Adjacent sequences: A277959 A277960 A277961 * A277963 A277964 A277965

Keyword

nonn

Author

Michel Lagneau, Nov 06 2016

Status

approved