|
|
A066773
|
|
Smallest number which requires n^2 steps in the 3x+1 problem.
|
|
2
|
|
|
1, 2, 16, 12, 7, 98, 153, 169, 673, 350, 107, 129, 649, 2110, 4763, 6919, 6943, 158299, 71310, 724686, 845223, 665215, 2157291, 4468260, 5978623, 34385063, 21015006, 301695657, 489853918, 568097511, 418606034, 1208474114
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
The next term is >= 1410123943. - Larry Reeves (larryr(AT)acm.org), Apr 10 2002
|
|
LINKS
|
|
|
EXAMPLE
|
a(3)=12 since the Collatz sequence is 12 -> 6 -> 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 and the count of steps is 9.
|
|
MATHEMATICA
|
f[ n_ ] := Module[ {i = 0, m = n}, While[ m != 1, m = If[ OddQ[ m ], 3m + 1, m/2 ]; i++ ]; i ]; a = Table[ 0, {50} ]; Do[ m = Sqrt[ f[ n ] ]; If[ IntegerQ[ m ] && a[ [ m + 1 ] ] == 0, a[ [ m + 1 ] ] = n ] ], {n, 1, 10^6} ]; a
|
|
PROG
|
(PARI) {sequence(n)= c=0; k=n; while(k>1, if(k%2==0, k=k/2, k=3*k+1); c=c+1; ); if(issquare(c), print(n, " ", c), ); }
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
More terms from Larry Reeves (larryr(AT)acm.org), Apr 10 2002
|
|
STATUS
|
approved
|
|
|
|