Base | Representation |
---|---|
bin | 101101110001011111011100… |
… | …101010011110101100101001 |
3 | 222101210100100001202122110011 |
4 | 231301133130222132230221 |
5 | 202341303224222112441 |
6 | 1552053545025051521 |
7 | 60255251432314222 |
oct | 5561373452365451 |
9 | 871710301678404 |
10 | 201313114254121 |
11 | 59165459857791 |
12 | 1a6b39b6786ba1 |
13 | 884398c31c7ac |
14 | 379d86cb9a249 |
15 | 1841939484581 |
hex | b717dca9eb29 |
201313114254121 has 2 divisors, whose sum is σ = 201313114254122. Its totient is φ = 201313114254120.
The previous prime is 201313114254103. The next prime is 201313114254157. The reversal of 201313114254121 is 121452411313102.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 199677812947600 + 1635301306521 = 14130740^2 + 1278789^2 .
It is a cyclic number.
It is not a de Polignac number, because 201313114254121 - 247 = 60575625898793 is a prime.
It is not a weakly prime, because it can be changed into another prime (201313114254521) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 100656557127060 + 100656557127061.
It is an arithmetic number, because the mean of its divisors is an integer number (100656557127061).
Almost surely, 2201313114254121 is an apocalyptic number.
It is an amenable number.
201313114254121 is a deficient number, since it is larger than the sum of its proper divisors (1).
201313114254121 is an equidigital number, since it uses as much as digits as its factorization.
201313114254121 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5760, while the sum is 31.
Adding to 201313114254121 its reverse (121452411313102), we get a palindrome (322765525567223).
The spelling of 201313114254121 in words is "two hundred one trillion, three hundred thirteen billion, one hundred fourteen million, two hundred fifty-four thousand, one hundred twenty-one".
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