14113113245041 has 4 divisors (see below), whose sum is σ = 14114904939252. Its totient is φ = 14111321550832.

The previous prime is 14113113245039. The next prime is 14113113245047. The reversal of 14113113245041 is 14054231131141.

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 870683074816 + 13242430170225 = 933104^2 + 3639015^2 .

It is a cyclic number.

It is not a de Polignac number, because 14113113245041 - 2¹ = 14113113245039 is a prime.

It is a super-2 number, since 2×14113113245041² (a number of 27 digits) contains 22 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 14113113244994 and 14113113245012.

It is not an unprimeable number, because it can be changed into a prime (14113113245047) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 895835290 + ... + 895851043.

It is an arithmetic number, because the mean of its divisors is an integer number (3528726234813).

Almost surely, 2^{14113113245041} is an apocalyptic number.

It is an amenable number.

14113113245041 is a deficient number, since it is larger than the sum of its proper divisors (1791694211).

14113113245041 is an equidigital number, since it uses as much as digits as its factorization.

14113113245041 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 1791694210.

The product of its (nonzero) digits is 5760, while the sum is 31.

Adding to 14113113245041 its reverse (14054231131141), we get a palindrome (28167344376182).

The spelling of 14113113245041 in words is "fourteen trillion, one hundred thirteen billion, one hundred thirteen million, two hundred forty-five thousand, forty-one".