Base | Representation |
---|---|
bin | 11101000111111101111… |
… | …111110000010011111001 |
3 | 21002100000110222022020212 |
4 | 131013331333300103321 |
5 | 230242402303004431 |
6 | 4131235025000505 |
7 | 264412040002004 |
oct | 35077577602371 |
9 | 7070013868225 |
10 | 2001421141241 |
11 | 70188606a285 |
12 | 283a7ba83135 |
13 | 11696bb05852 |
14 | 6cc2550653b |
15 | 370dcb4b72b |
hex | 1d1fdff04f9 |
2001421141241 has 2 divisors, whose sum is σ = 2001421141242. Its totient is φ = 2001421141240.
The previous prime is 2001421141217. The next prime is 2001421141243. The reversal of 2001421141241 is 1421411241002.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1947308611600 + 54112529641 = 1395460^2 + 232621^2 .
It is a cyclic number.
It is not a de Polignac number, because 2001421141241 - 210 = 2001421140217 is a prime.
Together with 2001421141243, 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 (2001421141243) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1000710570620 + 1000710570621.
It is an arithmetic number, because the mean of its divisors is an integer number (1000710570621).
Almost surely, 22001421141241 is an apocalyptic number.
It is an amenable number.
2001421141241 is a deficient number, since it is larger than the sum of its proper divisors (1).
2001421141241 is an equidigital number, since it uses as much as digits as its factorization.
2001421141241 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 512, while the sum is 23.
Adding to 2001421141241 its reverse (1421411241002), we get a palindrome (3422832382243).
The spelling of 2001421141241 in words is "two trillion, one billion, four hundred twenty-one million, one hundred forty-one thousand, two hundred forty-one".
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