[QUOTE=science_man_88;248936]how I interpret this is :

[CODE](09:53)>v=vector(10,n,if(n>1,2*v[n-1]+1,1))
%13 = [1, 3, 7, 15, 31, 63, 127, 255, 511, 1023]
(09:53)>for(n=1,#v,v[n]=v[n]^1/n)
(09:53)>v
%14 = [1, 3/2, 7/3, 15/4, 31/5, 21/2, 127/7, 255/8, 511/9, 1023/10]
(09:53)>for(n=1,#v,v[n]=v[n]-3/2)
(09:53)>v
%15 = [-1/2, 0, 5/6, 9/4, 47/10, 9, 233/14, 243/8, 995/18, 504/5][/CODE]

as you can see in the last v not all conform to v[n]<3/2, which blows the statement you supposedly lay your whole conjecture on to high heaven.even placing parentheses correctly I still get it false for both mersenne numbers and mersenne primes.[/QUOTE]

Try just making a list.

N MPE[sub]N[/sub] MPE[sub]N[/sub]^(1.500000),

and notice the fluctuating approach to and then around
the value 3/2.

[QUOTE=davar55;249211]
N MPE[sub]N[/sub] MPE[sub]N[/sub]^(1.500000),[/QUOTE]

[TEX]n\times MPE_N \times MPE_N^{3/2}[/TEX]

this is how I interpret that which means you are saying that a [TEX]\text {integer>=1}* \text {integer>1}* \text {integer>1}^{3/2} = 3/2 [/TEX] ?

? lets take the lowest case of 2 1*2*2^(1/5) which is around 5.65 if we say it has to be 1.5 you're way off are you saying that [TEX]\text {that answer} \div {MPE_{N+1}} = 3/2[/TEX] then I'm more likely to believe it using the first example which would bring it to about 1.89

Now be careful sm, don't assume that what davar writes is actually correct, even once you remove the obvious mistakes.

[QUOTE=CRGreathouse;249227]Now be careful sm, don't assume that what davar writes is actually correct, even once you remove the obvious mistakes.[/QUOTE]

if a mistake is obvious to me it's unlikely that the thing holds any actual facts.

[QUOTE=science_man_88;249230]if a mistake is obvious to me it's unlikely that the thing holds any actual facts.[/QUOTE]

Fair enough. You'll also note the lack of evidence -- the standard conjecture has better numerical evidence as well as a strong heuristic.

best i can do is get it to 7 through possible alteration. then it fails I've tried a number spiral for the last digits of all integers >=1 and then circled the digit where the exponents would land I have a few ideas but I don't have a concrete way to present them, maybe a spread sheet.

[QUOTE=davar55;249211]Try just making a list.

N MPE[sub]N[/sub] MPE[sub]N[/sub]^(1.500000),

and notice the fluctuating approach to and then around
the value 3/2.[/QUOTE]

Once again for me, having to hurry due to limited
computing time resources (though I'm no longer
riding a dinosaur), that post by me was wrong.
You really should think first before "blindly" following
anyone, including me.

The YJ-Conjecture limit formula is:

lim(n->infinity) M[sub]n[/sub] ^ (1/n) = 3/2 = 1.500000.

So the list I'm suggesting you compute is:

n, MPE[sub]n[/sub], MPE[sub]n[/sub]^(1.0 / n)

for n = 1 to 47

where MPE[sub]n[/sub] is the list

2,3,5,7,13,17,19,31,61,89,107,127,521,607, etc.

Then we can analyze the data I started with in 2006.

[QUOTE=davar55;249352]Once again for me, having to hurry due to limited
computing time resources (though I'm no longer
riding a dinosaur), that post by me was wrong.
You really should think first before "blindly" following
anyone, including me.

The YJ-Conjecture limit formula is:

lim(n->infinity) M[sub]n[/sub] ^ (1/n) = 3/2 = 1.500000.

So the list I'm suggesting you compute is:

n, MPE[sub]n[/sub], MPE[sub]n[/sub]^(1.0 / n)

for n = 1 to 47

where MPE[sub]n[/sub] is the list

2,3,5,7,13,17,19,31,61,89,107,127,521,607, etc.

Then we can analyze the data I started with in 2006.[/QUOTE]

[CODE](17:02)>for(i=1,#mersenne,print(mersenne[i]^(1.0/i)))
2.000000000000000000000000000
1.732050807568877293527446342
1.709975946676696989353108873
1.626576561697785743211232345
1.670277652334810394803652891
1.603521621512549282985464791
1.522926998218753393386968192
1.536102555746382080084096039
1.578955831901959581920696844
1.566531025721843309228146507
1.529288540910522460366821574
1.497328106146992604557789778
1.618033530221347008876089003
1.580517075546684774482807483
1.611108181063782432675943740
1.617850569292629664391427757
1.575929486505463156125518121
1.566240292511125441127408419
1.552329702833776473429342949
1.521548663732084805037571386
1.548184829102329494595765216
1.519502318204047920019162208
1.499943367067368910338914811
1.510610475500344547104696107
1.490936647375361341678740151
1.472006957225681423614603740
1.486484542508562273885684721
1.500637696861403754501242636
1.492483197974171054909171632
1.481464232274212955846795928
1.486226683055510708608763914
1.526577508382904969438978292
1.512950108495140761550758073
1.511472357027693313061657786
1.498263869087911248387534693
1.512948108348191840117822958
1.496720384272055787678110414
1.513872584170716147906235743
1.523357195557165263568423205
1.524240229323290374602268512[/CODE]

[CODE](17:06)>plot(i=1,#mersenne,mersenne[floor(i)]^(1/floor(i)),1.4,2)

2 ""'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''|
|: |
|: |
|: |
|: |
| : |
| : |
| : |
| : |
| __ |
| x |
| : |
| : "" |
| xx __ _ |
| x : "" |
| : xx__ : " xx__ |
| : _ : " x |
| xx " : xx xx _ "___ __ __xx
| "" "" xx _""x____ "" x |
| "" |
| |
1.4 |..............................................................|
1 40[/CODE]

with the equation he just posted it now becomes clear that he's pointing towards [TEX]\sqrt[n] {MPE_n}[/TEX] tends towards 1.5

[CODE]1.5 2 1.13 1.33
2.25 3 1.13 1.33
3.38 5 1.01 1.48
5.06 7 1.08 1.38
7.59 13 0.88 1.71
11.39 17 1.01 1.49
17.09 19 1.35 1.11
25.63 31 1.24 1.21
38.44 61 0.95 1.59
57.67 89 0.97 1.54
86.5 107 1.21 1.24
129.75 127 1.53 0.98
194.62 521 0.56 2.68
291.93 607 0.72 2.08
437.89 1279 0.51 2.92
656.84 2203 0.45 3.35
985.26 2281 0.65 2.32
1477.89 3217 0.69 2.18
2216.84 4253 0.78 1.92
3325.26 4423 1.13 1.33
4987.89 9689 0.77 1.94
7481.83 9941 1.13 1.33
11222.74 11213 1.5 1
16834.11 19937 1.27 1.18
25251.17 21701 1.75 0.86
37876.75 23209 2.45 0.61
56815.13 44497 1.92 0.78
85222.69 86243 1.48 1.01
127834.04 110503 1.74 0.86
191751.06 132049 2.18 0.69
287626.59 216091 2 0.75
431439.88 756839 0.86 1.75
647159.82 859433 1.13 1.33
970739.74 1257787 1.16 1.3
1456109.61 1398269 1.56 0.96
2184164.41 2976221 1.1 1.36
3276246.61 3021377 1.63 0.92
4914369.92 6972593 1.06 1.42
7371554.88 13466917 0.82 1.83
11057332.32 20996011 0.79 1.9
[/CODE]


this the first is the calculated value if my interpretation is correct the second is the actual value the third is (estimated / actual) ? The fourth is (actual/estimated) ? anyone see anything useful ?

[QUOTE=science_man_88;249373][CODE]1.5 2 1.13 1.33
2.25 3 1.13 1.33
3.38 5 1.01 1.48
5.06 7 1.08 1.38
7.59 13 0.88 1.71
11.39 17 1.01 1.49
17.09 19 1.35 1.11
25.63 31 1.24 1.21
38.44 61 0.95 1.59
57.67 89 0.97 1.54
86.5 107 1.21 1.24
129.75 127 1.53 0.98
194.62 521 0.56 2.68
291.93 607 0.72 2.08
437.89 1279 0.51 2.92
656.84 2203 0.45 3.35
985.26 2281 0.65 2.32
1477.89 3217 0.69 2.18
2216.84 4253 0.78 1.92
3325.26 4423 1.13 1.33
4987.89 9689 0.77 1.94
7481.83 9941 1.13 1.33
11222.74 11213 1.5 1
16834.11 19937 1.27 1.18
25251.17 21701 1.75 0.86
37876.75 23209 2.45 0.61
56815.13 44497 1.92 0.78
85222.69 86243 1.48 1.01
127834.04 110503 1.74 0.86
191751.06 132049 2.18 0.69
287626.59 216091 2 0.75
431439.88 756839 0.86 1.75
647159.82 859433 1.13 1.33
970739.74 1257787 1.16 1.3
1456109.61 1398269 1.56 0.96
2184164.41 2976221 1.1 1.36
3276246.61 3021377 1.63 0.92
4914369.92 6972593 1.06 1.42
7371554.88 13466917 0.82 1.83
11057332.32 20996011 0.79 1.9
[/CODE]


this the first is the calculated value if my interpretation is correct the second is the actual value the third is (estimated[COLOR="Red"](n+1)[/COLOR] / actual) ? The fourth is (actual/estimated) ? anyone see anything useful ?[/QUOTE]
correction