13443401250445 has 4 divisors (see below), whose sum is σ = 16132081500540. Its totient is φ = 10754721000352.

The previous prime is 13443401250361. The next prime is 13443401250479. The reversal of 13443401250445 is 54405210434431.

13443401250445 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in 2 ways, for example, as 10735675053156 + 2707726197289 = 3276534^2 + 1645517^2 .

It is a cyclic number.

It is not a de Polignac number, because 13443401250445 - 2³³ = 13434811315853 is a prime.

It is a super-2 number, since 2×13443401250445² (a number of 27 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 13443401250398 and 13443401250407.

It is a congruent number.

It is an unprimeable number.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1344340125040 + ... + 1344340125049.

It is an arithmetic number, because the mean of its divisors is an integer number (4033020375135).

Almost surely, 2^{13443401250445} is an apocalyptic number.

It is an amenable number.

13443401250445 is a deficient number, since it is larger than the sum of its proper divisors (2688680250095).

13443401250445 is an equidigital number, since it uses as much as digits as its factorization.

13443401250445 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 2688680250094.

The product of its (nonzero) digits is 460800, while the sum is 40.

Adding to 13443401250445 its reverse (54405210434431), we get a palindrome (67848611684876).

The spelling of 13443401250445 in words is "thirteen trillion, four hundred forty-three billion, four hundred one million, two hundred fifty thousand, four hundred forty-five".