4331234501401 has 4 divisors (see below), whose sum is σ = 4332122600292. Its totient is φ = 4330346402512.

The previous prime is 4331234501371. The next prime is 4331234501431. The reversal of 4331234501401 is 1041054321334.

4331234501401 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 is an interprime number because it is at equal distance from previous prime (4331234501371) and next prime (4331234501431).

It can be written as a sum of positive squares in 2 ways, for example, as 1355126481801 + 2976108019600 = 1164099^2 + 1725140^2 .

It is a cyclic number.

It is not a de Polignac number, because 4331234501401 - 2¹⁵ = 4331234468633 is a prime.

It is a super-2 number, since 2×4331234501401² (a number of 26 digits) contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (4331234501431) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 444042130 + ... + 444051883.

It is an arithmetic number, because the mean of its divisors is an integer number (1083030650073).

Almost surely, 2^{4331234501401} is an apocalyptic number.

It is an amenable number.

4331234501401 is a deficient number, since it is larger than the sum of its proper divisors (888098891).

4331234501401 is an equidigital number, since it uses as much as digits as its factorization.

4331234501401 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 888098890.

The product of its (nonzero) digits is 17280, while the sum is 31.

Adding to 4331234501401 its reverse (1041054321334), we get a palindrome (5372288822735).

The spelling of 4331234501401 in words is "four trillion, three hundred thirty-one billion, two hundred thirty-four million, five hundred one thousand, four hundred one".