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Jokodo:
Thanks, I'll look into this using base 6....
instead of 9 I'll use 5 when halving:
5 = 5
2.3 = 2+3 = 5
1.13 = 1+1+3 = 5
0.343 = 3+4+3 = 14 = 1+4 = 5
0.1513 = 1+5+1+3 = 14 = 1+4 = 5
Note I did things like (5/16).toString(6) and (3+4+3).toString(6)
I think 5 is more of a magic number than 9.... see the pentagram in post 18
Code:
function digitRoot(num, base) {
return ((parseInt(num, base) - 1) % (base - 1)) + 1;
}
And now about the Fibonacci sequence:
0: 1 = 1 [1]
1: 1 = 1 [1]
2: 2 = 2 [2]
3: 3 = 3 [3]
4: 5 = 5 [5]
5: 8 = 12 [3]
6: 13 = 21 [3]
7: 21 = 33 [1]
8: 34 = 54 [4]
9: 55 = 131 [5]
10: 89 = 225 [4]
11: 144 = 400 [4]
12: 233 = 1025 [3]
13: 377 = 1425 [2]
14: 610 = 2454 [5]
15: 987 = 4323 [2]
16: 1597 = 11221 [2]
17: 2584 = 15544 [4]
18: 4181 = 31205 [1]
19: 6765 = 51153 [5]
20: 10946 = 122402 [1]
21: 17711 = 213555 [1]
22: 28657 = 340401 [2]
23: 46368 = 554400 [3]
24: 75025 = 1335201 [5]
25: 121393 = 2334001 [3]
26: 196418 = 4113202 [3]
27: 317811 = 10451203 [1]
28: 514229 = 15004405 [4]
29: 832040 = 25500012 [5]
30: 1346269 = 44504421 [4]
#0 to #9
1,1,2,3,5,3,3,1,4,5
#10 to #19
4,4,3,2,5,2,2,4,1,5
that involves a pair of numbers, then two other numbers that add up to 5, then a 5...
I think base 6 is better since the pattern repeats faster. Note that in base 6, 5+5 = 14 = 5....
I guess the digital root sum of the digital roots is the same as the digital roots of the regular sum
doubling:
1 = 1
2 = 2
4 = 4
8 = 12 = 3
16 = 24 = 1
32 = 52 = 2
64 = 144 = 4
128 = 332 = 3
The pattern in base 6:
View attachment 24290
