An integer sequence puzzle

[CraigD]

Consider the integer sequence beginning with these 512 integers: 1,2,2,2,3,3,3,2,3,4,5,3,4,4,4,2,3,4,5,4,6,6,3,3,5,5,6,4,5,5,5,2,3,4,5,4,6,6,7,4,5,7,7,6,8,4,4,3,5,6,5,5,8,7,7,4,6,6,7,5,6,6,6,2,3,4,5,4,6,6,7,4,6,7,9,6,7,8,5,4,6,6,8,7,7,8,7,6,9,9,5,4,6,5,5,3,5,6,6,6,9,6,7,5,6,9,8,7,8,8,4,4,7,7,8,6,9,8,8,5,7,7,8,6,7,7,7,2,3,4,5,4,6,6,7,4,6,7,9,6,8,8,9,4,5,7,7,7,9,10,6,6,9,8,10,8,5,6,6,4,6,7,9,6,8,9,9,7,9,8,8,8,9,8,9,6,8,10,9,9,13,6,6,4,7,7,6,5,7,6,6,3,5,6,6,6,9,7,8,6,9,10,8,6,6,8,8,5,7,7,10,9,8,9,7,7,11,9,10,8,10,5,5,4,7,8,7,7,9,9,7,6,6,10,11,8,10,9,9,5,8,8,9,7,10,9,9,6,8,8,9,7,8,8,8,2,3,4,5,4,6,6,7,4,6,7,9,6,8,8,9,4,6,7,9,7,10,10,5,6,8,9,9,8,9,10,6,4,6,6,8,7,10,8,10,7,7,10,10,10,9,7,8,6,9,10,7,8,11,11,10,8,11,6,8,6,6,7,7,4,6,7,9,7,9,10,8,6,9,9,12,9,8,10,9,7,10,10,8,8,9,9,9,8,9,10,10,8,11,10,7,6,9,9,9,10,11,10,12,9,10,14,7,6,6,7,6,4,8,8,8,7,11,7,7,5,8,8,7,6,8,7,7,3,5,6,6,6,9,7,8,6,9,10,9,7,10,9,10,6,7,10,10,10,7,9,7,6,9,7,10,8,8,9,6,5,8,8,10,7,9,11,11,9,7,9,8,9,11,8,9,7,9,12,9,9,12,11,9,8,12,11,12,5,7,6,6,4,7,8,8,8,8,8,8,7,11,10,12,9,10,8,8,6,8,7,11,10,11,12,8,8,11,11,10,9,10,10,5,5,9,9,10,8,11,10,11,7,8,11,12,9,11,10,10,6,9,9,10,8,11,10,10,7,9,9,10,8,9,9,9,2

 

The puzzle is simple: find an expression that generates the sequence.

 

There’s only one rule: the answer can’t be a degree 511 polynomial that generates these 512 values. You’re welcome to post this answer just to show off your (or your computer’s :shrug:) polynomial fitting skills, but it won’t count as a win.

 

Hints:

• The sequence isn’t (to the best of my knowledge) generatable by a simple polynomial
  
• I generated it using
  • A generalization of the coefficients of the expected values in this post
    
  • Something involving base 2 (binary) numbers
    

  [*]If you know the answer, you’d almost certainly know how to generate the sequence that begins with the following 512 numbers:

  1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,5,6,6,7,6,7,7,8,6,7,7,8,7,8,8,9,1

[modest]

Frustrating. [edit: In a good way :)]

 

I figured out the second sequence, but I can't figure out a way to make an expression of it. For the integer 10:

 

A(2⁰)+B(2¹)+C(2²)+D(2³)

 

I realize what A, B, C, and D represent as it relates to binary numbers (that A and C should be zero while B and D are one), but I cannot figure out how to make an expression realize it.

 

I would think A somehow represents that 10/2 has a remainder of zero, B that 5/2 has r=1, C that 2/2 has r=0, and D that 1/2 has r=1.

 

That's a guess, and I'm stuck. :hihi:

 

~modest

[CraigD]

Here’s an example of a sequence generated using only the first 100 integers and the “something involving base 2 (binary) numbers” part of the algorithm used to generate the puzzle sequence:

1

1

2 1

2 2 3 1

2 2 3 2 3 3 4 1

2 2 3 2 3 3 4 2 3 3 4 3 4 4 5 1

2 2 3 2 3 3 4 2 3 3 4 3 4 4 5 2 3 3 4 3 4 4 5 3 4 4 5 4 5 5 6 1

2 2 3 2 3 3 4 2 3 3 4 3 4 4 5 2 3 3 4 3 4 4 5 3 4 4 5 4 5 5 6 2 3 3 4 3

[modest]

Could you verify that I have interpreted your second sequence correctly? (I dare not say: usefully) That is, summing the digits in a binary count

base 10 = binary => binary sum = digit in sequence

1

= 1 => 1 =

1

2

= 10 => 1+0 =

1

3

= 11 => 1+1 =

2

4

= 100 => 1+0+0 =

1

5

= 101 => 1+0+1 =

2

6

= 110 => 1+1+0 =

2

7

= 111 => 1+1+1 =

3

8

= 1000 => 1+0+0+0 =

1

9

= 1001 => 1+0+0+1 =

2

10

= 1010 => 1+0+1+0 =

2

gives the sequence 1,1,2,1,2,2,3,1,2,2... For which you say “If you know the answer, you’d almost certainly know how to generate the sequence that begins with the following 512 numbers:”. If this is indeed the way the second sequence was meant to be interpreted then you could perhaps give a nod.

 

~modest

[CraigD]

Could you verify that I have interpreted your second sequence correctly?

Yes, you have.

 

I’m honestly not sure if anyone short of an intuitive mathematical genius (not that I’m suggesting modest, or any other member, isn’t :)) could guess the next parts of the puzzle without my post #1 clue linking to this post, so modest or any other puzzler, please don't feel restrained in followin it, unless you're committed to an excercise in intuition, or some formal method beyond my comprehension.

 

In the interest of not excessively teasing at the puzzle, I’ll throw out another clue: rational numbers.

[Turtle]

Yes, you have. :thumbs_up

 

I’m honestly not sure if anyone short of an intuitive mathematical genius (not that I’m suggesting modest, or any other member, isn’t :)) could guess the next parts of the puzzle without my post #1 clue linking to this post, so modest or any other puzzler, please don't feel restrained in followin it, unless you're committed to an excercise in intuition, or some formal method beyond my comprehension.

 

In the interest of not excessively teasing at the puzzle, I’ll throw out another clue: rational numbers.

 

i would never have seen that! :doh: still, it wasn't for a lack of looking. :D anyway, i keep looking at your 'clue' post & either i'm missing something even more obvious or you have made an error. to whit:

For example, the expected value of a fair 6-sided die roll is

[math]1\frac16+2\frac16+3\frac16+4\frac16+5\frac16+6\frac16 = \frac{21}6 = 3.5[/math]

which can be confirmed by noting that, after many die roles, the average roll approaches 3.5

 

since [math]1\frac{1}6[/math] = [math]\frac{7}6[/math] and [math]2\frac{1}6[/math] = [math]\frac{13}6[/math], aren't we already up to [math]\frac{20}6[/math] and with the 4 other terms aren't we at [math]\frac{95}6[/math] ? :confused:

[CraigD]

.. anyway, i keep looking at your 'clue' post & either i'm missing something even more obvious or you have made an error. to whit:

For example, the expected value of a fair 6-sided die roll is

[math]1\frac16+2\frac16+3\frac16+4\frac16+5\frac16+6\frac16 = \frac{21}6 = 3.5[/math]

which can be confirmed by noting that, after many die roles, the average roll approaches 3.5.

since [math]1\frac{1}6[/math] = [math]\frac{7}6[/math] and [math]2\frac{1}6[/math] = [math]\frac{13}6[/math], aren't we already up to [math]\frac{20}6[/math] and with the 4 other terms aren't we at [math]\frac{95}6[/math] ? :confused:

I believe you’re missing the traditional algebraic implied multiplication operator, Turtle.

 

That is

[math]a\frac{b}{c}[/math]

means

[math]a \cdot \frac{b}{c}[/math]

, not

[math]a +\frac{b}{c}[/math]

, as you’re reading it.

[Turtle]

I believe you’re missing the traditional algebraic implied multiplication operator, Turtle.

 

That is

[math]a\frac{b}{c}[/math]

means

[math]a \cdot \frac{b}{c}[/math]

, not

[math]a +\frac{b}{c}[/math]

, as you’re reading it.

 

:doh: roger. i'm back on it like an implied operator. ;)

[Turtle]

computer was off, i was in bed, & on the verge of sleep, but something about Craig making his list the first 512 elements kept nagging me. so 512 is [math]2^9[/MATH] and in binary that's 1000000000. so, every power of 2 the sequence gets back to one and then repeats the last pattern until every element is used, then takes that same pattern again but adds 1 to every element. each new group contains [math]2^n-2^{n-1}[/MATH] more elements than the last. (mathspeak for every group is twice as big as the last.;)) the separated line below in the code box is my addition of the next group, but it's not complete, but it's late & i'm outa steam.

 

 

1,
1,2,
1,2,2,3,
1,2,2,3,2,3,3,4,
1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,
1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,
1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,5,6,6,7,6,7,7,8,6,7,7,8,7,8,8,9,

1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8,5,6,6,7,6,7,7,8,6,7,7,8,7,8,8,9,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3, 4,4,5,4,5,5,6,4...

[Turtle]

yes? no? maybe? ;)

[CraigD]

As I hinted, modest surmised a couple of days ago, and I believe Turtle’s suggesting, a step in how I generated the sequences in post #1 was to sum the digits of the members of another integer sequence. This step of the puzzle solved, we can rewrite the sequences (for consistency’s sake, their first 512 terms):

 

1, 5, 5, 17, 13, 37, 25, 65, 41, 101, 61, 145, 85, 197, 113, 257, 145, 325, 181, 401, 221, 485, 265, 577, 313, 677, 365, 785, 421, 901, 481, 1025, 545, 1157, 613, 1297, 685, 1445, 761, 1601, 841, 1765, 925, 1937, 1013, 2117, 1105, 2305, 1201, 2501, 1301, 2705, 1405, 2917, 1513, 3137, 1625, 3365, 1741, 3601, 1861, 3845, 1985, 4097, 2113, 4357, 2245, 4625, 2381, 4901, 2521, 5185, 2665, 5477, 2813, 5777, 2965, 6085, 3121, 6401, 3281, 6725, 3445, 7057, 3613, 7397, 3785, 7745, 3961, 8101, 4141, 8465, 4325, 8837, 4513, 9217, 4705, 9605, 4901, 10001, 5101, 10405, 5305, 10817, 5513, 11237, 5725, 11665, 5941, 12101, 6161, 12545, 6385, 12997, 6613, 13457, 6845, 13925, 7081, 14401, 7321, 14885, 7565, 15377, 7813, 15877, 8065, 16385, 8321, 16901, 8581, 17425, 8845, 17957, 9113, 18497, 9385, 19045, 9661, 19601, 9941, 20165, 10225, 20737, 10513, 21317, 10805, 21905, 11101, 22501, 11401, 23105, 11705, 23717, 12013, 24337, 12325, 24965, 12641, 25601, 12961, 26245, 13285, 26897, 13613, 27557, 13945, 28225, 14281, 28901, 14621, 29585, 14965, 30277, 15313, 30977, 15665, 31685, 16021, 32401, 16381, 33125, 16745, 33857, 17113, 34597, 17485, 35345, 17861, 36101, 18241, 36865, 18625, 37637, 19013, 38417, 19405, 39205, 19801, 40001, 20201, 40805, 20605, 41617, 21013, 42437, 21425, 43265, 21841, 44101, 22261, 44945, 22685, 45797, 23113, 46657, 23545, 47525, 23981, 48401, 24421, 49285, 24865, 50177, 25313, 51077, 25765, 51985, 26221, 52901, 26681, 53825, 27145, 54757, 27613, 55697, 28085, 56645, 28561, 57601, 29041, 58565, 29525, 59537, 30013, 60517, 30505, 61505, 31001, 62501, 31501, 63505, 32005, 64517, 32513, 65537, 33025, 66565, 33541, 67601, 34061, 68645, 34585, 69697, 35113, 70757, 35645, 71825, 36181, 72901, 36721, 73985, 37265, 75077, 37813, 76177, 38365, 77285, 38921, 78401, 39481, 79525, 40045, 80657, 40613, 81797, 41185, 82945, 41761, 84101, 42341, 85265, 42925, 86437, 43513, 87617, 44105, 88805, 44701, 90001, 45301, 91205, 45905, 92417, 46513, 93637, 47125, 94865, 47741, 96101, 48361, 97345, 48985, 98597, 49613, 99857, 50245, 101125, 50881, 102401, 51521, 103685, 52165, 104977, 52813, 106277, 53465, 107585, 54121, 108901, 54781, 110225, 55445, 111557, 56113, 112897, 56785, 114245, 57461, 115601, 58141, 116965, 58825, 118337, 59513, 119717, 60205, 121105, 60901, 122501, 61601, 123905, 62305, 125317, 63013, 126737, 63725, 128165, 64441, 129601, 65161, 131045, 65885, 132497, 66613, 133957, 67345, 135425, 68081, 136901, 68821, 138385, 69565, 139877, 70313, 141377, 71065, 142885, 71821, 144401, 72581, 145925, 73345, 147457, 74113, 148997, 74885, 150545, 75661, 152101, 76441, 153665, 77225, 155237, 78013, 156817, 78805, 158405, 79601, 160001, 80401, 161605, 81205, 163217, 82013, 164837, 82825, 166465, 83641, 168101, 84461, 169745, 85285, 171397, 86113, 173057, 86945, 174725, 87781, 176401, 88621, 178085, 89465, 179777, 90313, 181477, 91165, 183185, 92021, 184901, 92881, 186625, 93745, 188357, 94613, 190097, 95485, 191845, 96361, 193601, 97241, 195365, 98125, 197137, 99013, 198917, 99905, 200705, 100801, 202501, 101701, 204305, 102605, 206117, 103513, 207937, 104425, 209765, 105341, 211601, 106261, 213445, 107185, 215297, 108113, 217157, 109045, 219025, 109981, 220901, 110921, 222785, 111865, 224677, 112813, 226577, 113765, 228485, 114721, 230401, 115681, 232325, 116645, 234257, 117613, 236197, 118585, 238145, 119561, 240101, 120541, 242065, 121525, 244037, 122513, 246017, 123505, 248005, 124501, 250001, 125501, 252005, 126505, 254017, 127513, 256037, 128525, 258065, 129541, 260101, 130561, 262145

 

and

 

1, 4, 3, 8, 5, 12, 7, 16, 9, 20, 11, 24, 13, 28, 15, 32, 17, 36, 19, 40, 21, 44, 23, 48, 25, 52, 27, 56, 29, 60, 31, 64, 33, 68, 35, 72, 37, 76, 39, 80, 41, 84, 43, 88, 45, 92, 47, 96, 49, 100, 51, 104, 53, 108, 55, 112, 57, 116, 59, 120, 61, 124, 63, 128, 65, 132, 67, 136, 69, 140, 71, 144, 73, 148, 75, 152, 77, 156, 79, 160, 81, 164, 83, 168, 85, 172, 87, 176, 89, 180, 91, 184, 93, 188, 95, 192, 97, 196, 99, 200, 101, 204, 103, 208, 105, 212, 107, 216, 109, 220, 111, 224, 113, 228, 115, 232, 117, 236, 119, 240, 121, 244, 123, 248, 125, 252, 127, 256, 129, 260, 131, 264, 133, 268, 135, 272, 137, 276, 139, 280, 141, 284, 143, 288, 145, 292, 147, 296, 149, 300, 151, 304, 153, 308, 155, 312, 157, 316, 159, 320, 161, 324, 163, 328, 165, 332, 167, 336, 169, 340, 171, 344, 173, 348, 175, 352, 177, 356, 179, 360, 181, 364, 183, 368, 185, 372, 187, 376, 189, 380, 191, 384, 193, 388, 195, 392, 197, 396, 199, 400, 201, 404, 203, 408, 205, 412, 207, 416, 209, 420, 211, 424, 213, 428, 215, 432, 217, 436, 219, 440, 221, 444, 223, 448, 225, 452, 227, 456, 229, 460, 231, 464, 233, 468, 235, 472, 237, 476, 239, 480, 241, 484, 243, 488, 245, 492, 247, 496, 249, 500, 251, 504, 253, 508, 255, 512, 257, 516, 259, 520, 261, 524, 263, 528, 265, 532, 267, 536, 269, 540, 271, 544, 273, 548, 275, 552, 277, 556, 279, 560, 281, 564, 283, 568, 285, 572, 287, 576, 289, 580, 291, 584, 293, 588, 295, 592, 297, 596, 299, 600, 301, 604, 303, 608, 305, 612, 307, 616, 309, 620, 311, 624, 313, 628, 315, 632, 317, 636, 319, 640, 321, 644, 323, 648, 325, 652, 327, 656, 329, 660, 331, 664, 333, 668, 335, 672, 337, 676, 339, 680, 341, 684, 343, 688, 345, 692, 347, 696, 349, 700, 351, 704, 353, 708, 355, 712, 357, 716, 359, 720, 361, 724, 363, 728, 365, 732, 367, 736, 369, 740, 371, 744, 373, 748, 375, 752, 377, 756, 379, 760, 381, 764, 383, 768, 385, 772, 387, 776, 389, 780, 391, 784, 393, 788, 395, 792, 397, 796, 399, 800, 401, 804, 403, 808, 405, 812, 407, 816, 409, 820, 411, 824, 413, 828, 415, 832, 417, 836, 419, 840, 421, 844, 423, 848, 425, 852, 427, 856, 429, 860, 431, 864, 433, 868, 435, 872, 437, 876, 439, 880, 441, 884, 443, 888, 445, 892, 447, 896, 449, 900, 451, 904, 453, 908, 455, 912, 457, 916, 459, 920, 461, 924, 463, 928, 465, 932, 467, 936, 469, 940, 471, 944, 473, 948, 475, 952, 477, 956, 479, 960, 481, 964, 483, 968, 485, 972, 487, 976, 489, 980, 491, 984, 493, 988, 495, 992, 497, 996, 499, 1000, 501, 1004, 503, 1008, 505, 1012, 507, 1016, 509, 1020, 511, 1024

 

Recall my “rational number” hint, and add to it an additional one – “irreducible fraction”. With them, and pairs of numbers, you should know what to do next. From there, with a key bit of preceding trickery, a common function fitting technique like Newton’s forward difference method, can find the final piece of the puzzle. (Additional clue: you don’t need nearly as many terms as I’ve included in these posts to solve the puzzle).

 

Of course, this all seems obvious to me, because I already know how I generated the original sequences. I’ll leave it to legitimate puzzle solvers who don’t already know its answer. :confused:

[Turtle]

As I hinted, modest surmised a couple of days ago, and I believe Turtle’s suggesting, a step in how I generated the sequences in post #1 was to sum the digits of the members of another integer sequence. This step of the puzzle solved, we can rewrite the sequences (for consistency’s sake, their first 512 terms):

 

[ lists removed for brevity's sake ;) ]

 

Recall my “rational number” hint, and add to it an additional one – “irreducible fraction”. With them, and pairs of numbers, you should know what to do next. From there, with a key bit of preceding trickery, a common function fitting technique like Newton’s forward difference method, can find the final piece of the puzzle. (Additional clue: you don’t need nearly as many terms as I’ve included in these posts to solve the puzzle).

 

Of course, this all seems obvious to me, because I already know how I generated the original sequences. I’ll leave it to legitimate puzzle solvers who don’t already know its answer. ;)

 

that is a lot like the joke about the ceiling to me; over my head. :hyper: i'll keep trying to find an intuitive opening though. i have the world's only i-stick and i know how to use it.

 

on an aside from earlier, your new link on irreducible fraction stubs to "vulgar fraction" and there i find a strong case for my earlier mistake/misunderstanding of your equation in Latex. what is standard in one context may not be so in another. adding multiplier symbols to your equation would not be incorrect and would remove all doubt as to what is intended.

 

Mixed numbers

A mixed number is the sum of a whole number and a proper fraction. This sum is implied without the use of any visible operator such as "+"; for example, in referring to two entire cakes and three quarters of another cake, the whole and fractional parts of the number are written next to each other: [math]2+\frac34=2\frac34[/math] ...

[CraigD]

The “generalization of this post” I mentioned in post #1 sprang from noticing that, in 19926 thread, the original post sets up the paradox with a description stating “one box has 3 times more than the other”, while the the wikipedia article “two envelopes problem” sets it up with one stating “one envelope contains twice [(2 times)] as much as the other”.

 

This leads, in the “find the expected value” step of the paradox setup, to different results: [math]\frac{5}{4}x[/math] for 2 times, [math]\frac{5}{3}x[/math] for 3.

 

Consider what the expected value is for any “contains N times as much” integer. You’ll get this rational number sequence:

[math]\frac{1}{1},\, \frac{5}{4},\, \frac{5}{3},\, \frac{17}{8},\, \frac{13}{5},\, \frac{37}{12},\, \frac{25}{7} \, \dots \, \frac{N^2}{2}+\frac{1}{2N}[/math]

 

My puzzle sequence is simply the bit counts of the numerators of that sequence, the second “clue” sequence the denominatiors.

 

I though these were some pretty sequence, and a tricky puzzle for a couple of reasons:

 

The sequences increase on average, but not monotonically. The sequences of the bitcounts has periodic 2 or 1 terms, while the numerator and denominator sequences alternate between high and low values.

 

Given the sequence [math]\frac{1}{1},\, \frac{5}{4},\, \frac{5}{3},\, \frac{17}{8},\, \frac{13}{5},\, \frac{37}{12},\, \frac{25}{7} \, \dots \, [/math], the “workhorse” methods of finding a polynomial that fits it, [math]\frac{N^2}{2}+\frac{1}{2N}[/math], forward differences (specifically, I usually use Lagrange’s method, though there are others), doesn’t work. The differences never become zero. If you change it by multiplying each term by its position in the sequence to get this one

[math]\frac{1}{1},\, \frac{5}{2},\, \frac{5}{1},\, \frac{17}{2},\, \frac{13}{1},\, \frac{37}{2},\, \frac{25}{1} \,\dots[/math]

the difference method quickly finds the polynomial [math]\frac{N^3}{2}+\frac12[/math].

 

This last trick – multiplying each term of a sequence by its position [math]N[/math] (or [math]N^m[/math]) seems a pretty valuable one in “find the polynomial that generated this sequence” type problems, as, if you figure out how to select the right [math]m[/math] it allows the difference method to succeed for polynomials with negative and/or non-integer exponents, like this puzzle's

[math]\frac{N^2}{2}+\frac{1}{2N}[/math]

, or, I think, more spectacular ones like

[math]N^2 +\sqrt{N} +\frac{1}{\sqrt{N}} +\frac{1}{N}[/math]

.