430122113 has 4 divisors (see below), whose sum is σ = 444953940. Its totient is φ = 415290288.

The previous prime is 430122079. The next prime is 430122127. The reversal of 430122113 is 311221034.

It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 311221034 = 2 ⋅155610517.

It can be written as a sum of positive squares in 2 ways, for example, as 337346689 + 92775424 = 18367^2 + 9632^2 .

It is a cyclic number.

It is not a de Polignac number, because 430122113 - 2¹⁰ = 430121089 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 430122091 and 430122100.

It is not an unprimeable number, because it can be changed into a prime (430122193) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (11) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 7415870 + ... + 7415927.

It is an arithmetic number, because the mean of its divisors is an integer number (111238485).

Almost surely, 2^430122113 is an apocalyptic number.

It is an amenable number.

430122113 is a deficient number, since it is larger than the sum of its proper divisors (14831827).

430122113 is a wasteful number, since it uses less digits than its factorization.

430122113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 14831826.

The product of its (nonzero) digits is 144, while the sum is 17.

The square root of 430122113 is about 20739.3855502037. The cubic root of 430122113 is about 754.8556734779.

Adding to 430122113 its reverse (311221034), we get a palindrome (741343147).

The spelling of 430122113 in words is "four hundred thirty million, one hundred twenty-two thousand, one hundred thirteen".