[seqfan] New sequence

DAN_CYN_J dan_cyn_j at comcast.net
Tue Apr 30 01:06:13 CEST 2013



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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