130043411433 has 24 divisors (see below), whose sum is σ = 202391614152. Its totient is φ = 79986417408.

The previous prime is 130043411419. The next prime is 130043411447. The reversal of 130043411433 is 334114340031.

130043411433 is a `hidden beast` number, since 1 + 300 + 4 + 341 + 14 + 3 + 3 = 666.

It is an interprime number because it is at equal distance from previous prime (130043411419) and next prime (130043411447).

It can be written as a sum of positive squares in 4 ways, for example, as 10943879769 + 119099531664 = 104613^2 + 345108^2 .

It is not a de Polignac number, because 130043411433 - 2⁴ = 130043411417 is a prime.

It is a super-2 number, since 2×130043411433² (a number of 23 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 130043411397 and 130043411406.

It is not an unprimeable number, because it can be changed into a prime (130043411473) by changing a digit.

It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 45666 + ... + 512027.

It is an arithmetic number, because the mean of its divisors is an integer number (8432983923).

Almost surely, 2^130043411433 is an apocalyptic number.

130043411433 is a gapful number since it is divisible by the number (13) formed by its first and last digit.

It is an amenable number.

130043411433 is a deficient number, since it is larger than the sum of its proper divisors (72348202719).

130043411433 is a wasteful number, since it uses less digits than its factorization.

130043411433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 559705 (or 559702 counting only the distinct ones).

The product of its (nonzero) digits is 5184, while the sum is 27.

Adding to 130043411433 its reverse (334114340031), we get a palindrome (464157751464).

The spelling of 130043411433 in words is "one hundred thirty billion, forty-three million, four hundred eleven thousand, four hundred thirty-three".