A277365 - OEIS

%I #13 Sep 14 2017 03:59:14

%S 2,6,12,20,34,49,56,72,98,112,144,176,196,228,224,272,344,406,392,384,

%T 448,520,576,688,688,772,913,912,912,1028,992,1040,1220,1152,1376,

%U 1624,1624,1708,1624,1728,1728,1824,2160,2080,2080,2215,2559,2752,2884,2884,2752

%N a(n) is the smallest number k such that f(k) = f(n) + f(n+1) and g(k) = g(n) + g(n+1), where f(n) (resp. g(n)) is the number of halving (resp. tripling) steps to reach 1 in the Collatz ('3x+1') problem.

%C A006666: Number of halving steps to reach 1 in '3x+1' problem.

%C A006667: Number of tripling steps to reach 1 in '3x+1' problem.

%C We observe an interesting property: the subsequence {b(i)} of perfect squares is {49, 144, 196, 576, 3844, 12544, 15376, 51529, 61504, 246016, ...} with the property that b(3) = 4*b(1), b(4) = 4*b(2), b(7) = 4*b(5), b(9) = 4*b(7), b(10) = 4*b(9), ...

%C The primes of the sequence are 2, 2953, 3739, 9931, 38303, 44641, ...

%e a(3) = 12 because (A006666(3), A006667(3)) = (f(3), g(3)) = (5, 2) => f(12) = f(3) + f(4) = 5 + 2 = 7 and g(12) = g(3) + g(4) = 2 + 0 = 2.

%p nn:=10^6:U:=array(1..nn):V:=array(1..nn):

%p for i from 1 to nn do:

%p m:=i:it0:=0:it1:=0:

%p    for j from 1 to nn while(m<>1) do:

%p     if irem(m,2)=0

%p      then

%p      m:=m/2:it0:=it0+1:

%p      else

%p      m:=3*m+1:it1:=it1+1:

%p     fi:

%p    od:

%p    U[i]:=it0:V[i]:=it1:

%p   od:

%p    for n from 1 to 100 do:

%p    ii:=0:

%p     for k from 1 to nn while(ii=0) do:

%p      if U[k]=U[n]+U[n+1] and V[k]=V[n]+V[n+1]

%p       then

%p       ii:=1:printf(`%d, `,k):

%p       else

%p      fi:

%p     od:

%p od:

%Y Cf. A006666, A006667.

%K nonn

%O 1,1

%A _Michel Lagneau_, Oct 11 2016

%E Name edited by _Michel Marcus_, Sep 13 2017