Base | Representation |
---|---|
bin | 11101010011001000000… |
… | …100011100000100001001 |
3 | 21010110221101202111021002 |
4 | 131103020010130010021 |
5 | 230441421213120012 |
6 | 4140535505224345 |
7 | 265314650454143 |
oct | 35231004340411 |
9 | 7113841674232 |
10 | 2013401301257 |
11 | 7069735aa875 |
12 | 2862640320b5 |
13 | 117b2ab11a28 |
14 | 6d64063bc93 |
15 | 3758e79d7c2 |
hex | 1d4c811c109 |
2013401301257 has 2 divisors, whose sum is σ = 2013401301258. Its totient is φ = 2013401301256.
The previous prime is 2013401301217. The next prime is 2013401301269. The reversal of 2013401301257 is 7521031043102.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1444313625616 + 569087675641 = 1201796^2 + 754379^2 .
It is a cyclic number.
It is not a de Polignac number, because 2013401301257 - 216 = 2013401235721 is a prime.
It is not a weakly prime, because it can be changed into another prime (2013401301217) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1006700650628 + 1006700650629.
It is an arithmetic number, because the mean of its divisors is an integer number (1006700650629).
Almost surely, 22013401301257 is an apocalyptic number.
It is an amenable number.
2013401301257 is a deficient number, since it is larger than the sum of its proper divisors (1).
2013401301257 is an equidigital number, since it uses as much as digits as its factorization.
2013401301257 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 5040, while the sum is 29.
Adding to 2013401301257 its reverse (7521031043102), we get a palindrome (9534432344359).
The spelling of 2013401301257 in words is "two trillion, thirteen billion, four hundred one million, three hundred one thousand, two hundred fifty-seven".
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