Base | Representation |
---|---|
bin | 111100100010101000… |
… | …0111111001101101101 |
3 | 110102120200211011222102 |
4 | 1321011100333031231 |
5 | 4112230320200023 |
6 | 135420513024445 |
7 | 12251540054624 |
oct | 1710520771555 |
9 | 412520734872 |
10 | 130011100013 |
11 | 501569707a3 |
12 | 21244555125 |
13 | c34c28cca2 |
14 | 6414b586bb |
15 | 35add07628 |
hex | 1e4543f36d |
130011100013 has 2 divisors, whose sum is σ = 130011100014. Its totient is φ = 130011100012.
The previous prime is 130011100003. The next prime is 130011100031. The reversal of 130011100013 is 310001110031.
It is a happy number.
Together with next prime (130011100031) it forms an Ormiston pair, because they use the same digits, order apart.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 80155801924 + 49855298089 = 283118^2 + 223283^2 .
It is an emirp because it is prime and its reverse (310001110031) is a distict prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-130011100013 is a prime.
It is a super-2 number, since 2×1300111000132 (a number of 23 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 130011099967 and 130011100003.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (130011100003) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65005550006 + 65005550007.
It is an arithmetic number, because the mean of its divisors is an integer number (65005550007).
Almost surely, 2130011100013 is an apocalyptic number.
It is an amenable number.
130011100013 is a deficient number, since it is larger than the sum of its proper divisors (1).
130011100013 is an equidigital number, since it uses as much as digits as its factorization.
130011100013 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 9, while the sum is 11.
Adding to 130011100013 its reverse (310001110031), we get a palindrome (440012210044).
The spelling of 130011100013 in words is "one hundred thirty billion, eleven million, one hundred thousand, thirteen".
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