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Title: "A week long problem"


Shaun - February 24, 2006 09:37 AM (GMT)
Endi remarked, in the "factorials" thread: This is where this site should be going - proper mathematics.

I went on a week-long Maths course once, with the Open University, staying at Stirling University.
Apart from the (much higher) Maths we were studying the tutors set various problems to keep us busy should we actually manage somehow to run out of work. The one presented in this thread was one of the "Week long problems".
At the end of the week we discovered the tutors didn't in fact have any answers for any of these problems ...

Here's the one I liked best.

Given a number, N.
If it's even, halve it.
If it's odd, multiply it by three, add 1, and then halve it.
Stop when you reach 1.

EG: starting with N=3
3
(3x3 +1 = 10)
5
(3x5 +1 = 16)
8
4
2
1

It takes 5 steps to reach 1: 5, 8, 4, 2, 1
let S(N) be the number of steps to reach 1 from N
S(3) = 5


EG N=9
9
(28)
14
7
(22)
11
(34)
17
(52)
26
13
(40)
20
10
5
(16)
8
4
2
1
9: 14 7 11 17 26 13 20 10 5 8 4 2 1
S(9) = 13

S(3) = 5; S(9) = 13.

Question: Can we get S(N) from N without writing out all the terms?
Is there a formula?

Dan - February 24, 2006 10:01 PM (GMT)
Here's a bunch more S values to work with:

S(1) = 0
S(2) = 1
S(3) = 5
S(4) = 2
S(5) = 4
S(6) = 6
S(7) = 11
S(8) = 3
S(9) = 13
S(10) = 5
S(11) = 10
S(12) = 7
S(13) = 7
S(14) = 12
S(15) = 12
S(16) = 4
S(17) = 9
S(18) = 14
S(19) = 14
S(20) = 6
S(21) = 6
S(22) = 11
S(23) = 11
S(24) = 8
S(25) = 16
S(26) = 8
S(27) = 70
S(28) = 13
S(29) = 13
S(30) = 13
S(31) = 67
S(32) = 5
S(33) = 18
S(34) = 10
S(35) = 10
S(36) = 15
S(37) = 15
S(38) = 15
S(39) = 23
S(40) = 7
S(41) = 69
S(42) = 7
S(43) = 20
S(44) = 12
S(45) = 12
S(46) = 12
S(47) = 66
S(48) = 9
S(49) = 17
S(50) = 17
S(51) = 17
S(52) = 9
S(53) = 9
S(54) = 71
S(55) = 71
S(56) = 14
S(57) = 22
S(58) = 14
S(59) = 22
S(60) = 14
S(61) = 14
S(62) = 68
S(63) = 68
S(64) = 6
S(65) = 19
S(66) = 19
S(67) = 19
S(68) = 11
S(69) = 11
S(70) = 11
S(71) = 65
S(72) = 16
S(73) = 73
S(74) = 16
S(75) = 11
S(76) = 16
S(77) = 16
S(78) = 24
S(79) = 24
S(80) = 8
S(81) = 16
S(82) = 70
S(83) = 70
S(84) = 8
S(85) = 8
S(86) = 21
S(87) = 21
S(88) = 13
S(89) = 21
S(90) = 13
S(91) = 59
S(92) = 13
S(93) = 13
S(94) = 67
S(95) = 67
S(96) = 10
S(97) = 75
S(98) = 18
S(99) = 18
S(100) = 18
S(101) = 18
S(102) = 18
S(103) = 56
S(104) = 10
S(105) = 26
S(106) = 10
S(107) = 64
S(108) = 72
S(109) = 72
S(110) = 72
S(111) = 45
S(112) = 15
S(113) = 10
S(114) = 23
S(115) = 23
S(116) = 15
S(117) = 15
S(118) = 23
S(119) = 23
S(120) = 15
S(121) = 61
S(122) = 15
S(123) = 31
S(124) = 69
S(125) = 69
S(126) = 69
S(127) = 31
S(128) = 7
S(129) = 77
S(130) = 20
S(131) = 20
S(132) = 20
S(133) = 20
S(134) = 20
S(135) = 28
S(136) = 12
S(137) = 58
S(138) = 12
S(139) = 28
S(140) = 12
S(141) = 12
S(142) = 66
S(143) = 66
S(144) = 17

Dan - February 24, 2006 10:07 PM (GMT)
The one thing that's obvious is that S(2^N) = N.

S(1) = 0
S(2) = 1
S(4) = 2
S(8) = 3
S(16) = 4
S(32) = 5
S(64) = 6
S(128) = 7

Dan - February 24, 2006 10:59 PM (GMT)
In general, if N=A×B, where A is odd and B is a power of 2, then S( N ) = S( A ) + S( B ).

S(6) = S(3×2) = S(3) + S(2) = 5 + 1 = 6
S(10) = S(5×2) = S(5) + S(2) = 4 + 1 = 5
S(12) = S(3×4) = S(3) + S(4) = 5 + 2 = 7
S(14) = S(7×2) = S(7) + S(2) = 11 + 1 = 12
S(18) = S(9×2) = S(9) + S(2) = 13 + 1 = 14
S(20) = S(5×4) = S(5) + S(4) = 4 + 2 = 6
S(22) = S(11×2) = S(11) + S(2) = 10 + 1 = 11
S(24) = S(3×8) = S(3) + S(8) = 5 + 3 = 8

Now all we need is a formula for when N is odd.

Shaun - February 25, 2006 04:52 PM (GMT)
QUOTE (Dan @ Feb 24 2006, 10:59 PM)
In general, if N=A×B, where A is odd and B is a power of 2, then S( N ) = S( A ) + S( B ).

Now all we need is a formula for when N is odd.

Yes, for any even number
S(k x 2^n) = S(k) + n.


Enigma - March 4, 2006 06:45 PM (GMT)
I found your problem - called the Syracuse algorithm - in the Penguin Book of Curious and Interesting Numbers, but the termsaren't counted in the same way.

For example, staring with 17 we have
17 52 26 13 40 20 10 5 16 8 4 2 1 with 12 steps, while you have
17 (52) 26 13 (40) 20 10 5 (16) 8 4 2 1 with 9 steps.

Counting the terms their way gives s(27) = 111 (reaching a maximum height of 9232) as against your s(27) = 70.

What do you, and Dan, think? Should you count all the terms produced as in the book?

Shaun - March 4, 2006 06:47 PM (GMT)
QUOTE (Enigma @ Mar 4 2006, 06:45 PM)
What do you, and Dan, think? Should you count all the terms produced as in the book?

I'll have another look at the problem and see if the book's ideas produce any interesting results.
My approach has got nowhere to date!

Dan - March 4, 2006 07:20 PM (GMT)
Mine hasn't gotten anywhere, either.

Computing S their way gives:

S(1) = 0
S(2) = 1
S(3) = 7
S(4) = 2
S(5) = 5
S(6) = 8
S(7) = 16
S(8) = 3
S(9) = 19
S(10) = 6
S(11) = 14
S(12) = 9
S(13) = 9
S(14) = 17
S(15) = 17
S(16) = 4
S(17) = 12
S(18) = 20
S(19) = 20
S(20) = 7
S(21) = 7
S(22) = 15
S(23) = 15
S(24) = 10
S(25) = 23
S(26) = 10
S(27) = 111
S(28) = 18
S(29) = 18
S(30) = 18
S(31) = 106
S(32) = 5
S(33) = 26
S(34) = 13
S(35) = 13
S(36) = 21
S(37) = 21
S(38) = 21
S(39) = 34
S(40) = 8
S(41) = 109
S(42) = 8
S(43) = 29
S(44) = 16
S(45) = 16
S(46) = 16
S(47) = 104
S(48) = 11
S(49) = 24
S(50) = 24
S(51) = 24
S(52) = 11
S(53) = 11
S(54) = 112
S(55) = 112
S(56) = 19
S(57) = 32
S(58) = 19
S(59) = 32
S(60) = 19
S(61) = 19
S(62) = 107
S(63) = 107
S(64) = 6
S(65) = 27
S(66) = 27
S(67) = 27
S(68) = 14
S(69) = 14
S(70) = 14
S(71) = 102
S(72) = 22
S(73) = 115
S(74) = 22
S(75) = 14
S(76) = 22
S(77) = 22
S(78) = 35
S(79) = 35
S(80) = 9
S(81) = 22
S(82) = 110
S(83) = 110
S(84) = 9
S(85) = 9
S(86) = 30
S(87) = 30
S(88) = 17
S(89) = 30
S(90) = 17
S(91) = 92
S(92) = 17
S(93) = 17
S(94) = 105
S(95) = 105
S(96) = 12
S(97) = 118
S(98) = 25
S(99) = 25
S(100) = 25
S(101) = 25
S(102) = 25
S(103) = 87
S(104) = 12
S(105) = 38
S(106) = 12
S(107) = 100
S(108) = 113
S(109) = 113
S(110) = 113
S(111) = 69
S(112) = 20
S(113) = 12
S(114) = 33
S(115) = 33
S(116) = 20
S(117) = 20
S(118) = 33
S(119) = 33
S(120) = 20
S(121) = 95
S(122) = 20
S(123) = 46
S(124) = 108
S(125) = 108
S(126) = 108
S(127) = 46
S(128) = 7
S(129) = 121
S(130) = 28
S(131) = 28
S(132) = 28
S(133) = 28
S(134) = 28
S(135) = 41
S(136) = 15
S(137) = 90
S(138) = 15
S(139) = 41
S(140) = 15
S(141) = 15
S(142) = 103
S(143) = 103
S(144) = 23

Shaun - March 4, 2006 07:30 PM (GMT)
"More work for the Undertaker" ...
old Music-hall song.
I'll have another look at it tomorrow.

Dan - March 5, 2006 07:52 AM (GMT)
QUOTE (Enigma @ Mar 4 2006, 12:45 PM)
I found your problem - called the Syracuse algorithm - in the Penguin Book of Curious and Interesting Numbers,

It's also in Kelly&Pohl's C by Dissection as the "hailstones" sequence. It says it's still an open question whether S(n) is even finite.

Ebbe - March 5, 2006 10:58 PM (GMT)
QUOTE
It says it's still an open question whether S(n) is even finite.


Yes, I thought I recognized it as this open problem too, but wasn't sure.
So whoever solves it, will get the praise of very many mathematicians... But I won't try, I've too often been sitting whole nights over (for me) unsolvable problems.

Dan - October 3, 2006 04:59 AM (GMT)
Potential relevance to base 12:

Suppose that (3n+1)/2 is even (and thus the n=even case is reached in the next step). Then 3n+1 is a multiple of 4. In dozenal, a number is a multiple of 4 iff it ends in 4, 8, or 0. Thus, 3n ends in 3 (because a multiple of 3 cannot end in 7 or :B). This means that the last digit of n is 1, 5, or 9.

To be continued...

Dan - October 3, 2006 05:59 AM (GMT)
If n is an odd number, then n mod 4 is either 1 (i.e., the numbers ending in 1, 5, or 9 mentioned above) or 3 (numbers ending in 3, 7, or :B in dozenal).

Case 1: n=4m+1
S(n) = S(4m+1)
= 1 + S(3(4m+1)+1)
= 1 + S(12m + 4)
= 2 + S(6m + 2)
= 3 + S(3m + 1)

Case 3: n=4m+3
S(n) = S(4m+3)
= 1 + S(3(4m+3)+1)
= 1 + S(12m + 10)
= 2 + S(6m + 5)
= 3 + S(3(6m+5) + 1)
= 3 + S(18m + 16)
= 4 + S(9m + 8)

In summary:
  • If n=2m (i.e., is even), then S(n) = 1+S(m)
  • If n=4m+1, then S(n)=3+S(3m+1)
  • If n=4m+3, then S(n)=4+S(9m+8)
Note that the first two cases express S(n) in terms of S(a number less than n), but the third does not. This case will be considered in my next post later.

Dan - October 23, 2006 10:09 PM (GMT)
Maybe a graph would be helpful.

user posted image

Hyperkubus - October 24, 2006 02:58 PM (GMT)
Oooh, a forum semi-about duodecimals.. Sweet!

Anyways, on topic, here's a graph from 0-400 000:
user posted image

It certainly looks like the distribution it has a distinct 'shape', so I tried to find a function for that shape. The first succes I got was this:
The lowest S(n) numbers all lie on the function log(n)/log(2) (as in the 2-log of n)

Then I tried to find other functions for the dots which didn't lie on this function. Since S(5)=5, I tried inserting a variable at places and calculating
a*log(5)/log(2)=4
log(5)/log(a)=4
log(a5)/log(2)=4

This last one gave me a proper function, running through the first dots of horizontal series of dots.
a prooved to be 16/5 making the formula
log(16n/5)/log(2)
or
log(n2^4/5)/log(2)

other functions I found were
log(n2^5/3)/log(2)
log(n2^9/17)/log(2)
log(n2^10/11)/log(2)
log(n2^10/17)/log(2)

This gives the graph:
user posted image

I don't know why the formulas are like this, I haven't proven it, and I haven't yet looked at the 'high' numbers (see Dan's graph) (I'm actually not even sure if there actually are 'high' numbers or if I can 'hit' them with my formula)

Maybe someone else can find a pattern here, going from what I've found..

Dan - October 24, 2006 04:06 PM (GMT)
QUOTE (Hyperkubus @ Oct 24 2006, 08:58 AM)
The first succes I got was this:
The lowest S(n) numbers all lie on the function log(n)/log(2) (as in the 2-log of n)

Those are the powers of two: S(2^m) = m

Dan - August 4, 2007 03:15 AM (GMT)
The following values of n=am+b allow S(n) to be written as S(cm+d)+e where c<a:

2m, 4m, 4m+1, 4m+2, 6m, 6m+2, 6m+4, 8m, 8m+2, 8m+4, 8m+5, 10m, 10m+2, 10m+4, 10m+6, 10m+8, 12m, 12m+1, 12m+2, 12m+4, 12m+5, 12m+6, 12m+8, 12m+9, 12m+10, 14m, 14m+2, 14m+4, 14m+6, 14m+8, 14m+10, 14m+12, 16m, 16m+1, 16m+2, 16m+3, 16m+4, 16m+5, 16m+6, 16m+8, 16m+10, 16m+12, 16m+13, 18m, 18m+2, 18m+4, 18m+6, 18m+8, 18m+10, 18m+12, 18m+14, 18m+16, 20m, 20m+1, 20m+2, 20m+4, 20m+5, 20m+6, 20m+8, 20m+9, 20m+10, 20m+12, 20m+13, 20m+14, 20m+16, 20m+17, 20m+18, 22m, 22m+2, 22m+4, 22m+6, 22m+8, 22m+10, 22m+12, 22m+14, 22m+16, 22m+18, 22m+20, 24m, 24m+2, 24m+4, 24m+5, 24m+8, 24m+10, 24m+12, 24m+13, 24m+16, 24m+18, 24m+20, 24m+21, 26m, 26m+2, 26m+4, 26m+6, 26m+8, 26m+10, 26m+12, 26m+14, 26m+16, 26m+18, 26m+20, 26m+22, 26m+24, 28m, 28m+1, 28m+2, 28m+4, 28m+5, 28m+6, 28m+8, 28m+9, 28m+10, 28m+12, 28m+13, 28m+14, 28m+16, 28m+17, 28m+18, 28m+20, 28m+21, 28m+22, 28m+24, 28m+25, 28m+26, 30m, 30m+2, 30m+4, 30m+6, 30m+8, 30m+10, 30m+12, 30m+14, 30m+16, 30m+18, 30m+20, 30m+22, 30m+24, 30m+26, 30m+28, 32m, 32m+1, 32m+2, 32m+3, 32m+4, 32m+5, 32m+6, 32m+8, 32m+10, 32m+11, 32m+12, 32m+13, 32m+14, 32m+16, 32m+17, 32m+18, 32m+19, 32m+20, 32m+21, 32m+22, 32m+23, 32m+24, 32m+25, 32m+26, 32m+28, 32m+29, 34m, 34m+2, 34m+4, 34m+6, 34m+8, 34m+10, 34m+12, 34m+14, 34m+16, 34m+18, 34m+20, 34m+22, 34m+24, 34m+26, 34m+28, 34m+30, 34m+32, 36m, 36m+1, 36m+2, 36m+4, 36m+5, 36m+6, 36m+8, 36m+9, 36m+10, 36m+12, 36m+13, 36m+14, 36m+16, 36m+17, 36m+18, 36m+20, 36m+21, 36m+22, 36m+24, 36m+25, 36m+26, 36m+28, 36m+29, 36m+30, 36m+32, 36m+33, 36m+34, 38m, 38m+2, 38m+4, 38m+6, 38m+8, 38m+10, 38m+12, 38m+14, 38m+16, 38m+18, 38m+20, 38m+22, 38m+24, 38m+26, 38m+28, 38m+30, 38m+32, 38m+34, 38m+36, 40m, 40m+2, 40m+4, 40m+5, 40m+8, 40m+10, 40m+12, 40m+13, 40m+16, 40m+18, 40m+20, 40m+21, 40m+24, 40m+26, 40m+28, 40m+29, 40m+32, 40m+34, 40m+36, 40m+37, 42m, 42m+2, 42m+4, 42m+6, 42m+8, 42m+10, 42m+12, 42m+14, 42m+16, 42m+18, 42m+20, 42m+22, 42m+24, 42m+26, 42m+28, 42m+30, 42m+32, 42m+34, 42m+36, 42m+38, 42m+40, 44m, 44m+1, 44m+2, 44m+4, 44m+5, 44m+6, 44m+8, 44m+9, 44m+10, 44m+12, 44m+13, 44m+14, 44m+16, 44m+17, 44m+18, 44m+20, 44m+21, 44m+22, 44m+24, 44m+25, 44m+26, 44m+28, 44m+29, 44m+30, 44m+32, 44m+33, 44m+34, 44m+36, 44m+37, 44m+38, 44m+40, 44m+41, 44m+42, 46m, 46m+2, 46m+4, 46m+6, 46m+8, 46m+10, 46m+12, 46m+14, 46m+16, 46m+18, 46m+20, 46m+22, 46m+24, 46m+26, 46m+28, 46m+30, 46m+32, 46m+34, 46m+36, 46m+38, 46m+40, 46m+42, 46m+44, 48m, 48m+1, 48m+2, 48m+3, 48m+4, 48m+5, 48m+6, 48m+8, 48m+10, 48m+12, 48m+13, 48m+16, 48m+17, 48m+18, 48m+19, 48m+20, 48m+21, 48m+22, 48m+24, 48m+26, 48m+28, 48m+29, 48m+32, 48m+33, 48m+34, 48m+35, 48m+36, 48m+37, 48m+38, 48m+40, 48m+42, 48m+44, 48m+45, 50m, 50m+2, 50m+4, 50m+6, 50m+8, 50m+10, 50m+12, 50m+14, 50m+16, 50m+18, 50m+20, 50m+22, 50m+24, 50m+26, 50m+28, 50m+30, 50m+32, 50m+34, 50m+36, 50m+38, 50m+40, 50m+42, 50m+44, 50m+46, 50m+48, 52m, 52m+1, 52m+2, 52m+4, 52m+5, 52m+6, 52m+8, 52m+9, 52m+10, 52m+12, 52m+13, 52m+14, 52m+16, 52m+17, 52m+18, 52m+20, 52m+21, 52m+22, 52m+24, 52m+25, 52m+26, 52m+28, 52m+29, 52m+30, 52m+32, 52m+33, 52m+34, 52m+36, 52m+37, 52m+38, 52m+40, 52m+41, 52m+42, 52m+44, 52m+45, 52m+46, 52m+48, 52m+49, 52m+50, 54m, 54m+2, 54m+4, 54m+6, 54m+8, 54m+10, 54m+12, 54m+14, 54m+16, 54m+18, 54m+20, 54m+22, 54m+24, 54m+26, 54m+28, 54m+30, 54m+32, 54m+34, 54m+36, 54m+38, 54m+40, 54m+42, 54m+44, 54m+46, 54m+48, 54m+50, 54m+52, 56m, 56m+2, 56m+4, 56m+5, 56m+8, 56m+10, 56m+12, 56m+13, 56m+16, 56m+18, 56m+20, 56m+21, 56m+24, 56m+26, 56m+28, 56m+29, 56m+32, 56m+34, 56m+36, 56m+37, 56m+40, 56m+42, 56m+44, 56m+45, 56m+48, 56m+50, 56m+52, 56m+53, 58m, 58m+2, 58m+4, 58m+6, 58m+8, 58m+10, 58m+12, 58m+14, 58m+16, 58m+18, 58m+20, 58m+22, 58m+24, 58m+26, 58m+28, 58m+30, 58m+32, 58m+34, 58m+36, 58m+38, 58m+40, 58m+42, 58m+44, 58m+46, 58m+48, 58m+50, 58m+52, 58m+54, 58m+56, 60m, 60m+1, 60m+2, 60m+4, 60m+5, 60m+6, 60m+8, 60m+9, 60m+10, 60m+12, 60m+13, 60m+14, 60m+16, 60m+17, 60m+18, 60m+20, 60m+21, 60m+22, 60m+24, 60m+25, 60m+26, 60m+28, 60m+29, 60m+30, 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