some analytics...
5598802 (2, 11, 254491)
degree factorization numbers, where among the numbers, there is our number 5598802
2^47395 1011100100100011 5, 9479
2^47853 1011101011101101 3, 3, 13, 409
2^48555 1011110110101011 3, 3, 5, 13, 83
2^48704 1011111001000000 2, 2, 2, 2, 2, 2, 761
2^49993 1100001101001001 499932^50806 1100011001110110 2, 7, 19, 191
2^50903 1100011011010111 109, 467
2^52592 1100110101110000 2, 2, 2, 2, 19, 173
2^53094 1100111101100110 2, 3, 8849
2^54821 1101011000100101 13, 4217
2^55112 1101011101001000 2, 2, 2, 83, 83
2^55655 1101100101100111 5, 11131
2^56022 1101101011010110 2, 3, 9337
2^56117 1101101100110101 17, 3301
2^57421 1110000001001101 7, 13, 631
2^57925 1110001001000101 5, 5, 7, 331
14428676 (2, 2, 19, 189851)
2^38669 1001011100001101 386692^42741 1010011011110101 3, 3, 3, 1583
2^43942 1010101110100110 2, 127, 173
2^52395 1100110010101011 3, 5, 7, 499
2^52461 1100110011101101 3, 3, 3, 29, 67
2^55196 1101011110011100 2, 2, 13799
2^58769 1110010110010001 17, 3457
2^63838 1111100101011110 2, 59, 541
what we see, that our number will be 100% in factorized numbers equal to themselves (and that there are just a lot of them among other factors))). there is also a small gain in the search 5598802 vs 15050 (2^49993,num length > 15050), 14428676 vs 11641 (2^38669,num length > 11641). although we can divide this length by the length of our number 14428676 > 8 11641:8= 1455,125 options (from left to right), total 14428676 vs 1456 winnings.
Themselves these numbers (equal to themselves) is also not a bit, from our 38669 to 49993 is 1060 steps.
2^[38669]
2^[38671]
2^[38677]
2^[38693]
2^[38699]
2^[38707]
2^[38711]
...
2^[49927]
2^[49937]
2^[49939]
2^[49943]
2^[49957]
2^[49991]
2^[49993]
In general, all this is necessary to digest, and try to catch something. Large numbers will have to be divided into pieces (the blessing the number we need floats from the beginning of a large number by the end) otherwise the pc will think for a long time. Or search, search algorithm. In theory, there should be numbers (in infinity) which will begin on the number of signs we need 18.19 .. pz59, pz60 .. what is their length, trillions of millions of characters...
script (peeped in google) factorization (integer "degrees") with the search for the desired number
def factors(n):
gaps = [1,2,2,4,2,4,2,4,6,2,6]
length, cycle = 11, 3
f, fs, nxt = 2, [], 0
while f * f <= n:
while n % f == 0:
fs.append(f)
n /= f
f += gaps[nxt]
nxt += 1
if nxt == length:
nxt = cycle
if n > 1: fs.append(n)
return fs
i = 38669 # 2^ start
while i <= 38670: # 2^ end
fac = str(factors(i))
ed = fac.count(",")
if ed == 0:
line = fac
bina = ''.join( c for c in line if c not in '[]' )
hhh = int(bina)
a = 2**hhh
n = str(a)
nn= str("14428676") # pz num
if nn in n:
print(fac,"< 2^ fac search","|","num length >",len(str(a))) #print("suchen",a)
i=i+1
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