How to Create a Spirallic Numbers Dataset

The Tutorial

1. How to Create a Spirallic Numbers Dataset Picture 1
2. Make the dataset by opening a new Excel worksheet and choosing a cell inwards and downwards about 10 rows and columns to start inputting the numbers into. You may start with 0 or 1 or any number; 1 is recommended.
3. Proceed sequentially outwards in a spiral. If you want to save a little time, use a formula like "=J15+1" and copy it up, across or down off as many cells as need be. You may then find a like formula nearby as you go outwards, which will speed things up greatly. This will also allow you to use any originating number in the center, should you so choose..
4. After you have built the data set up to 100 or 324 (as in the example), you're good to go!
5. You will notice right away that the prime numbers are fairly evenly distributed in the 4 quadrants at this point. You might try blocking in non-primes in groups in patterns to see if you can predict primes!
6. Notice there are a number of sequences to the columns and numbers which follow Summation Notation.
7. Notice also that the first sequence is easy -- it's the diagonal from the center to the bottom right. Its pattern to produce the sequence 1,3,13,31,57,91,133,183,241,307, ... is -1*0+1=1, 1*2+1=3, 3*4+1 =13, 5*6+1 = 31, etc. You will find a similar pattern extending to the upper left.

   1. Notice that the second pattern is trickier: the vertical pattern that produces the sequence 1,4,15,34,61,96,139,190,249,316 is more involved. If one takes the difference of those numbers and then the differences of the differences, one gets 8's, so one knows that a constant value of 8 is being added in the sequence. Knowing that, a typical rate of change is produced by ((a*b)+c)+8 = result, with a being constant. Such is the case here as it turns out, for the sequence is produced by ((4*-2)+4+8)=4, ((4*1)+3+8)=15, ((4*6)+2+8)=34, ((4*13)+1+8) = 61, etc. with the increment to b being 3,5,7,... and the increment to c being -1, while a=4 and 8 are constants. There's an elegant way to write that in Summation Notation with N's and K's and i's and all but it's beyond the text capacity we're operating under. Actually, I just found another way to resolve the series: 1==((-1*0)-7)+8, 4=((1*2)-6)+8, 15=((3*4)-5)+8, 34=((5*6)-4)+8, 61=((7*8)-3)+8, 96=((9*10)-2)+8, etc.
8. Notice the horizontal pattern is a bit trickier to discover. 8's were again a constant difference of the differences but it took me awhile to find the rest of the formula. The sequence of 11,28,53,86,127,176,233,298, ... is generated as far as I know by ((4*-5)+23)+8=11, ((4*-1)+24)+8 = 28, ((4*5)+25)+8 = 53, ((4*13)+26)+8 = 86, etc., with a=4 and 8 remaining constant while b increments by 4, 6, 8,... and c incrementing by 1. Again, there's a more elegant way to state that. If anyone happening to read this can help out, great. There's another way to resolve this sequence also: 2=((2*3)-12)+8, 11=((4*5)-17)+8, 28=((6*7)-22)+8, 53=((8*9)-27)+8, 86=((10*11)-32)+8, 127=((12*13)-37)+8, 176=((14*15)-42)+8, 233=((16*17)-47)+8, and 298 =((18*19)-52)+8, etc. with element c decrementing by -5 each time.
9. Notice that the diagonals take the form of ((a*b)+c)+d = Sum wherein b_sub_0 is 1 greater than a_sub_zero, a_sub_1 was 2 greater than a_sub_0 and b_sub_1 was 2 greater than b_sub_0, i.e. (1*2) changed next to (3*4). This might be the general pattern instead of working with 4's as had been, The role of d=8, the difference of the differences, was still present, so then you just need to figure out which a and b pair best approximated the answer and adjust with c accordingly, by trial and error. It did not take long to hit upon the solutions.
10. There are resolved 2 more sequences: 5=((2*3)-9)+8, 18=((4*5)-10)+8, 39=((6*7)-11)+8, etc. and also 8==((2*3)-6)+8, 23=((4*5)-5)+8, 46=((6*7)-4)+8, 77=((8*9)-3)+8, 116=((10*11)-2)+8,163=((12*13)-1)+8, 218=((14*15)-0)+8 and 281=((16*17)+1)+8. These were done by the analytical method described above in Step 9, though it's a good hunch the problem will only yield to greater finesse than the lazy mathematician possesses.