Base | Representation |
---|---|
bin | 1110111001101101111… |
… | …0010010010010110001 |
3 | 220110210212202120220202 |
4 | 3232123132102102301 |
5 | 13143302414300233 |
6 | 313335434244545 |
7 | 24332126442152 |
oct | 3563336222261 |
9 | 813725676822 |
10 | 256011478193 |
11 | 99634894849 |
12 | 41749886755 |
13 | 1b1ac646293 |
14 | c568d6b929 |
15 | 69d596cbe8 |
hex | 3b9b7924b1 |
256011478193 has 2 divisors, whose sum is σ = 256011478194. Its totient is φ = 256011478192.
The previous prime is 256011478153. The next prime is 256011478253. The reversal of 256011478193 is 391874110652.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 252333419584 + 3678058609 = 502328^2 + 60647^2 .
It is a cyclic number.
It is not a de Polignac number, because 256011478193 - 28 = 256011477937 is a prime.
It is not a weakly prime, because it can be changed into another prime (256011478103) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 128005739096 + 128005739097.
It is an arithmetic number, because the mean of its divisors is an integer number (128005739097).
It is a 1-persistent number, because it is pandigital, but 2⋅256011478193 = 512022956386 is not.
Almost surely, 2256011478193 is an apocalyptic number.
It is an amenable number.
256011478193 is a deficient number, since it is larger than the sum of its proper divisors (1).
256011478193 is an equidigital number, since it uses as much as digits as its factorization.
256011478193 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 362880, while the sum is 47.
The spelling of 256011478193 in words is "two hundred fifty-six billion, eleven million, four hundred seventy-eight thousand, one hundred ninety-three".
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