141112434000301 has 4 divisors (see below), whose sum is σ = 141538755553504. Its totient is φ = 140686112447100.

The previous prime is 141112434000299. The next prime is 141112434000353. The reversal of 141112434000301 is 103000434211141.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 141112434000301 - 2¹ = 141112434000299 is a prime.

It is a Duffinian number.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (141112434000371) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 213160776105 + ... + 213160776766.

It is an arithmetic number, because the mean of its divisors is an integer number (35384688888376).

Almost surely, 2^{141112434000301} is an apocalyptic number.

It is an amenable number.

141112434000301 is a deficient number, since it is larger than the sum of its proper divisors (426321553203).

141112434000301 is an equidigital number, since it uses as much as digits as its factorization.

141112434000301 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 426321553202.

The product of its (nonzero) digits is 1152, while the sum is 25.

Adding to 141112434000301 its reverse (103000434211141), we get a palindrome (244112868211442).

The spelling of 141112434000301 in words is "one hundred forty-one trillion, one hundred twelve billion, four hundred thirty-four million, three hundred one", and thus it is an aban number.