1411310131141 has 4 divisors (see below), whose sum is σ = 1411317124192. Its totient is φ = 1411303138092.

The previous prime is 1411310131111. The next prime is 1411310131193.

1411310131141 is nontrivially palindromic in base 10.

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 1411310131141 - 2¹³ = 1411310122949 is a prime.

It is a super-2 number, since 2×1411310131141² (a number of 25 digits) contains 22 as substring.

It is a Duffinian number.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (1411310131111) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 3184521 + ... + 3600526.

It is an arithmetic number, because the mean of its divisors is an integer number (352829281048).

Almost surely, 2^{1411310131141} is an apocalyptic number.

It is an amenable number.

1411310131141 is a deficient number, since it is larger than the sum of its proper divisors (6993051).

1411310131141 is an equidigital number, since it uses as much as digits as its factorization.

1411310131141 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 6993050.

The product of its (nonzero) digits is 144, while the sum is 22.

It can be divided in two parts, 1411310 and 131141, that added together give a palindrome (1542451).

The spelling of 1411310131141 in words is "one trillion, four hundred eleven billion, three hundred ten million, one hundred thirty-one thousand, one hundred forty-one".