14011341431130 has 48 divisors (see below), whose sum is σ = 36431422789320. Its totient is φ = 3736159246464.

The previous prime is 14011341431099. The next prime is 14011341431161. The reversal of 14011341431130 is 3113414311041.

14011341431130 is a `hidden beast` number, since 1 + 4 + 0 + 1 + 1 + 341 + 4 + 311 + 3 + 0 = 666.

It is an interprime number because it is at equal distance from previous prime (14011341431099) and next prime (14011341431161).

It can be written as a sum of positive squares in 4 ways, for example, as 13078226002689 + 933115428441 = 3616383^2 + 965979^2 .

It is a junction number, because it is equal to n+sod(n) for n = 14011341431094 and 14011341431103.

It is an unprimeable number.

It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 2427117 + ... + 5823536.

Almost surely, 2^{14011341431130} is an apocalyptic number.

14011341431130 is a gapful number since it is divisible by the number (10) formed by its first and last digit.

14011341431130 is an abundant number, since it is smaller than the sum of its proper divisors (22420081358190).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

14011341431130 is a wasteful number, since it uses less digits than its factorization.

14011341431130 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 8269535 (or 8269532 counting only the distinct ones).

The product of its (nonzero) digits is 1728, while the sum is 27.

Adding to 14011341431130 its reverse (3113414311041), we get a palindrome (17124755742171).

The spelling of 14011341431130 in words is "fourteen trillion, eleven billion, three hundred forty-one million, four hundred thirty-one thousand, one hundred thirty".