A277365 - OEIS

A277365

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.

2, 6, 12, 20, 34, 49, 56, 72, 98, 112, 144, 176, 196, 228, 224, 272, 344, 406, 392, 384, 448, 520, 576, 688, 688, 772, 913, 912, 912, 1028, 992, 1040, 1220, 1152, 1376, 1624, 1624, 1708, 1624, 1728, 1728, 1824, 2160, 2080, 2080, 2215, 2559, 2752, 2884, 2884, 2752

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OFFSET

1,1

COMMENTS

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

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

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), ...

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

LINKS

Table of n, a(n) for n=1..51.

EXAMPLE

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.

MAPLE

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

for i from 1 to nn do:

m:=i:it0:=0:it1:=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:it1:=it1+1:

fi:

od:

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

od:

for n from 1 to 100 do:

ii:=0:

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

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

then

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

else

fi:

od:

CROSSREFS

Cf. A006666, A006667.

Sequence in context: A266194 A194110 A291876 * A184432 A003274 A259470

Adjacent sequences: A277362 A277363 A277364 * A277366 A277367 A277368

KEYWORD

nonn

AUTHOR

Michel Lagneau, Oct 11 2016

EXTENSIONS

Name edited by Michel Marcus, Sep 13 2017

STATUS

approved