131044443350153 has 4 divisors (see below), whose sum is σ = 132839572711188. Its totient is φ = 129249313989120.

The previous prime is 131044443350147. The next prime is 131044443350239. The reversal of 131044443350153 is 351053344440131.

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 23346706776649 + 107697736573504 = 4831843^2 + 10377752^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-131044443350153 is a prime.

It is a super-2 number, since 2×131044443350153² (a number of 29 digits) contains 22 as substring.

It is a Duffinian number.

It is a Curzon number.

It is a self number, because there is not a number n which added to its sum of digits gives 131044443350153.

It is not an unprimeable number, because it can be changed into a prime (131044403350153) 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, 897564680408 + ... + 897564680553.

It is an arithmetic number, because the mean of its divisors is an integer number (33209893177797).

Almost surely, 2^{131044443350153} is an apocalyptic number.

It is an amenable number.

131044443350153 is a deficient number, since it is larger than the sum of its proper divisors (1795129361035).

131044443350153 is an equidigital number, since it uses as much as digits as its factorization.

131044443350153 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 1795129361034.

The product of its (nonzero) digits is 518400, while the sum is 41.

Adding to 131044443350153 its reverse (351053344440131), we get a palindrome (482097787790284).

The spelling of 131044443350153 in words is "one hundred thirty-one trillion, forty-four billion, four hundred forty-three million, three hundred fifty thousand, one hundred fifty-three".