[seqfan] Re: New sequence

[Robert G. Wilson v]

Where is the pair {1,2}? Bob.

-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of DAN_CYN_J
Sent: Monday, April 29, 2013 7:06 PM
To: seqfan at list.seqfan.eu
Subject: [seqfan] New sequence



Hi all seq. fans, 

  

A sequence of only primes that are closest to the next square > then itself. 

3,7,13,23,31,47,61,89,97,113,139,167,193,223,251,283,317,359,397,439,479... 

Not in OEIS 

2^n where n=
1,2,3,4,5... 

2^n sequence that is related term for term showing each pairs relation. Where the square is related to the first prime < then itself. 
4,9,16,25,36,49,64,91,100,121,144,169,196,225,256,289,324,361,400,441,484... 

The related pairs below. 
3,4
7,9
13,16
23,25
etc. 

What will add to the interest of the first sequence of primes is using the Ulam spiral (counter clockwise) in 3d and summing all primes starting @ (2) creating a prime tower. 
All primes < 4 (2*2) = 2,3 then  2+3=5
Consider each prime as 1 unit thick and laid out in Ulam spiral fashion and stacked one on top of each other giving the sum of primes. 
Then the number of primes (2) * (2*2) =8 Which for the first iteration only is a perfect cube 2*2*2 Then subtract 8-5 = 3 = the non prime space left in the cube. It has 5/3 ratio. 
It shifts for all future primes where the next cubical rectangle is 3*3*4 and producing 2+3+5+7 =17. 
4*9 =36 where (4) stands for the # of primes < 9 and 36 stands for the total cubicle units within the cubical rectangle. 
So in this case and all future cases there will possibly be more non prime space then prime space in these cubical rectangles. 
36-17 =
19 = non prime cubical space. 
17 = cubical prime space. 
19/17 = ratio. 
If using my first sequence of primes above --- 3,7,13,23,31,47,61,89,97,113,139,167,193,223,251,283,317,359,397,439,479... 
and still use all the primes in this rectangle stack, the ratio will eventually converge to(1) at the point of each prime in the above sequence. 
Where the non-prime space in the cubical rectangle will possibly be slightly > then the prime cubical space when this sequential stack is huge. These primes above will give the smallest possible ratio. 


To get a better perspective, in choosing prime 359 find the sum of all primes too 359 sum = 11599 Total number of primes =72 The next largest square = 19^2 = 361
72*361 = 25992 =total cubical number in the rectangular stack. 
25992 - prime sum(11599) = 14393 non prime space in the cubical rectangle. 
14393/11599 = ratio = 1.240882834727.. 

I conjecture that this ratio will converge to (1) as this sequence --->oo below. 
3,7,13,23,31,47,61,89,97,113... 

Also in choosing these special primes it gives the smallest ratio 

possible between the non prime cubical area and the prime 

cubical area for each succeeding square.    



Could the tipping point happen when the prime cubical area will ever be > then the non prime cubical area of the rectangle? 



By j ust entering in OEIS the special prime sequence probably is not interesting 
enough but adding the rest will peek the interest! 




Any questions are welcome! 

    
Dan 

  


 

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