141131132001241 has 4 divisors (see below), whose sum is σ = 147267268175232. Its totient is φ = 134994995827252.

The previous prime is 141131132001221. The next prime is 141131132001271. The reversal of 141131132001241 is 142100231131141.

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 141131132001241 - 2¹⁵ = 141131131968473 is a prime.

It is a super-3 number, since 3×141131132001241³ (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (141131132001221) 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, 3068068086961 + ... + 3068068087006.

It is an arithmetic number, because the mean of its divisors is an integer number (36816817043808).

Almost surely, 2^{141131132001241} is an apocalyptic number.

It is an amenable number.

141131132001241 is a deficient number, since it is larger than the sum of its proper divisors (6136136173991).

141131132001241 is an equidigital number, since it uses as much as digits as its factorization.

141131132001241 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 6136136173990.

The product of its (nonzero) digits is 576, while the sum is 25.

Adding to 141131132001241 its reverse (142100231131141), we get a palindrome (283231363132382).

The spelling of 141131132001241 in words is "one hundred forty-one trillion, one hundred thirty-one billion, one hundred thirty-two million, one thousand, two hundred forty-one".