Base | Representation |
---|---|
bin | 11101010010110110010… |
… | …110011110101000100001 |
3 | 21010110011200102002010201 |
4 | 131102312112132220201 |
5 | 230440314113243411 |
6 | 4140450210025201 |
7 | 265304411003512 |
oct | 35226626365041 |
9 | 7113150362121 |
10 | 2013104040481 |
11 | 706830827406 |
12 | 2861a0578201 |
13 | 117ab1362991 |
14 | 6d612d7ca09 |
15 | 375736313c1 |
hex | 1d4b659ea21 |
2013104040481 has 2 divisors, whose sum is σ = 2013104040482. Its totient is φ = 2013104040480.
The previous prime is 2013104040457. The next prime is 2013104040499. The reversal of 2013104040481 is 1840404013102.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1741080250000 + 272023790481 = 1319500^2 + 521559^2 .
It is a cyclic number.
It is not a de Polignac number, because 2013104040481 - 211 = 2013104038433 is a prime.
It is a super-2 number, since 2×20131040404812 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (2013104440481) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1006552020240 + 1006552020241.
It is an arithmetic number, because the mean of its divisors is an integer number (1006552020241).
Almost surely, 22013104040481 is an apocalyptic number.
It is an amenable number.
2013104040481 is a deficient number, since it is larger than the sum of its proper divisors (1).
2013104040481 is an equidigital number, since it uses as much as digits as its factorization.
2013104040481 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 3072, while the sum is 28.
Adding to 2013104040481 its reverse (1840404013102), we get a palindrome (3853508053583).
The spelling of 2013104040481 in words is "two trillion, thirteen billion, one hundred four million, forty thousand, four hundred eighty-one".
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