- FindStat

Identifier

St000826: Binary words ⟶ ℤ

Values

0 => 1

1 => 7

00 => 2

01 => 5

10 => 8

11 => 16

000 => 3

001 => 19

010 => 6

011 => 14

100 => 9

101 => 9

110 => 17

111 => 17

0000 => 4

0001 => 12

0010 => 20

0011 => 20

0100 => 7

0101 => 7

0110 => 15

0111 => 15

1000 => 10

1001 => 23

1010 => 10

1011 => 111

1100 => 18

1101 => 18

1110 => 18

1111 => 106

00000 => 5

00001 => 26

00010 => 13

00011 => 13

00100 => 21

00101 => 21

00110 => 21

00111 => 34

01000 => 8

01001 => 109

01010 => 8

01011 => 29

01100 => 16

01101 => 16

01110 => 16

01111 => 104

10000 => 11

10001 => 24

10010 => 24

10011 => 24

10100 => 11

10101 => 11

10110 => 112

10111 => 112

11000 => 19

11001 => 32

11010 => 19

11011 => 32

11100 => 19

11101 => 19

11110 => 107

11111 => 107

000000 => 6

000001 => 27

000010 => 27

000011 => 27

000100 => 14

000101 => 14

000110 => 14

000111 => 102

001000 => 22

001001 => 115

001010 => 22

001011 => 14

001100 => 22

001101 => 22

001110 => 35

001111 => 35

010000 => 9

010001 => 22

010010 => 110

010011 => 110

010100 => 9

010101 => 9

010110 => 30

010111 => 30

011000 => 17

011001 => 30

011010 => 17

011011 => 92

011100 => 17

011101 => 17

011110 => 105

011111 => 105

100000 => 12

100001 => 118

100010 => 25

100011 => 25

100100 => 25

100101 => 25

100110 => 25

>>> Load all 1200 entries. <<<

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$F_{1} = q + q^{7}$

$F_{2} = q^{2} + q^{5} + q^{8} + q^{16}$

$F_{3} = q^{3} + q^{6} + 2\ q^{9} + q^{14} + 2\ q^{17} + q^{19}$

$F_{4} = q^{4} + 2\ q^{7} + 2\ q^{10} + q^{12} + 2\ q^{15} + 3\ q^{18} + 2\ q^{20} + q^{23} + q^{106} + q^{111}$

$F_{5} = q^{5} + 2\ q^{8} + 3\ q^{11} + 2\ q^{13} + 3\ q^{16} + 4\ q^{19} + 3\ q^{21} + 3\ q^{24} + q^{26} + q^{29} + 2\ q^{32} + q^{34} + q^{104} + 2\ q^{107} + q^{109} + 2\ q^{112}$

$F_{6} = q^{6} + 3\ q^{9} + 4\ q^{12} + 4\ q^{14} + 4\ q^{17} + 5\ q^{20} + 5\ q^{22} + 5\ q^{25} + 3\ q^{27} + 3\ q^{30} + 4\ q^{33} + 2\ q^{35} + q^{38} + 2\ q^{46} + q^{69} + q^{87} + q^{92} + q^{95} + q^{100} + q^{102} + 2\ q^{105} + 3\ q^{108} + 2\ q^{110} + 3\ q^{113} + q^{115} + q^{118}$

$F_{7} = q^{7} + 3\ q^{10} + 6\ q^{13} + 6\ q^{15} + 6\ q^{18} + 7\ q^{21} + 7\ q^{23} + 8\ q^{26} + 5\ q^{28} + 6\ q^{31} + 6\ q^{34} + 4\ q^{36} + 4\ q^{39} + 2\ q^{41} + 3\ q^{44} + 5\ q^{47} + q^{49} + 3\ q^{52} + q^{54} + q^{65} + q^{67} + 2\ q^{70} + q^{80} + q^{83} + q^{85} + 2\ q^{88} + q^{90} + 2\ q^{93} + 2\ q^{96} + q^{98} + 2\ q^{101} + 2\ q^{103} + 3\ q^{106} + 4\ q^{109} + 3\ q^{111} + 4\ q^{114} + 3\ q^{116} + 4\ q^{119} + q^{121} + q^{124} + 2\ q^{127}$

$F_{8} = q^{8} + 4\ q^{11} + 7\ q^{14} + 8\ q^{16} + 10\ q^{19} + q^{21} + 10\ q^{22} + 11\ q^{24} + 14\ q^{27} + 9\ q^{29} + 11\ q^{32} + 11\ q^{35} + 8\ q^{37} + 10\ q^{40} + 7\ q^{42} + 9\ q^{45} + 10\ q^{48} + 4\ q^{50} + 8\ q^{53} + 3\ q^{55} + 3\ q^{58} + q^{60} + 2\ q^{61} + q^{63} + 2\ q^{66} + 2\ q^{68} + 3\ q^{71} + q^{73} + q^{76} + q^{78} + 2\ q^{81} + 2\ q^{84} + 2\ q^{86} + 3\ q^{89} + 2\ q^{91} + 3\ q^{94} + 5\ q^{97} + 2\ q^{99} + 3\ q^{102} + 3\ q^{104} + 4\ q^{107} + 5\ q^{110} + 5\ q^{112} + 6\ q^{115} + 6\ q^{117} + 8\ q^{120} + 4\ q^{122} + 5\ q^{125} + 6\ q^{128} + q^{130} + 3\ q^{133} + 2\ q^{141} + q^{143}$

$F_{9} = q^{9} + 4\ q^{12} + 9\ q^{15} + 10\ q^{17} + 15\ q^{20} + 3\ q^{22} + 14\ q^{23} + 18\ q^{25} + 25\ q^{28} + 20\ q^{30} + 21\ q^{33} + q^{35} + 20\ q^{36} + 15\ q^{38} + 21\ q^{41} + 15\ q^{43} + 19\ q^{46} + q^{48} + 18\ q^{49} + 10\ q^{51} + 14\ q^{54} + 8\ q^{56} + 10\ q^{59} + 3\ q^{61} + 5\ q^{62} + 7\ q^{64} + 6\ q^{67} + 5\ q^{69} + 6\ q^{72} + 3\ q^{74} + 3\ q^{77} + 2\ q^{79} + 3\ q^{82} + 5\ q^{85} + 4\ q^{87} + 5\ q^{90} + 3\ q^{92} + 2\ q^{93} + 6\ q^{95} + 9\ q^{98} + 4\ q^{100} + 5\ q^{103} + 4\ q^{105} + 5\ q^{108} + 7\ q^{111} + 7\ q^{113} + 11\ q^{116} + 10\ q^{118} + 13\ q^{121} + 9\ q^{123} + 11\ q^{126} + 15\ q^{129} + 7\ q^{131} + 9\ q^{134} + 2\ q^{136} + 3\ q^{139} + 6\ q^{142} + 4\ q^{144} + 5\ q^{147} + 2\ q^{152} + q^{155} + q^{170} + q^{173} + q^{178}$

Description

The stopping time of the decimal representation of the binary word for the 3x+1 problem.
Prepend $1$ to the binary word, interpret it as a positive number, and count the number of halving and tripling steps needed to reach $1$.

References

[1] wikipedia:3x_+_1_problem

Code

def statistic(w):
    return collatz(Integer("0b1" + str(w)))

@cached_function
def collatz(n):
    if n == 1:
        return 0
    elif is_even(n):
        return 1+collatz(n//2)
    else:
        return 1+collatz(3*n+1)

Created

May 30, 2017 at 22:00 by Martin Rubey

Updated

May 30, 2017 at 22:00 by Martin Rubey