Base | Representation |
---|---|
bin | 10010111011110000101… |
… | …100000100000001100011 |
3 | 11121101102011110012120002 |
4 | 102323300230010001203 |
5 | 132304142230043201 |
6 | 2433420410410215 |
7 | 163000631162336 |
oct | 22736054040143 |
9 | 4541364405502 |
10 | 1301118206051 |
11 | 46188a341a24 |
12 | 1901b9a3096b |
13 | 95905980b21 |
14 | 46d8dca661d |
15 | 23ca21a576b |
hex | 12ef0b04063 |
1301118206051 has 2 divisors, whose sum is σ = 1301118206052. Its totient is φ = 1301118206050.
The previous prime is 1301118206021. The next prime is 1301118206053. The reversal of 1301118206051 is 1506028111031.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1301118206051 - 218 = 1301117943907 is a prime.
It is a super-2 number, since 2×13011182060512 (a number of 25 digits) contains 22 as substring.
Together with 1301118206053, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (1301118206053) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 650559103025 + 650559103026.
It is an arithmetic number, because the mean of its divisors is an integer number (650559103026).
Almost surely, 21301118206051 is an apocalyptic number.
1301118206051 is a deficient number, since it is larger than the sum of its proper divisors (1).
1301118206051 is an equidigital number, since it uses as much as digits as its factorization.
1301118206051 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1440, while the sum is 29.
The spelling of 1301118206051 in words is "one trillion, three hundred one billion, one hundred eighteen million, two hundred six thousand, fifty-one".
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