11114311311121 has 2 divisors, whose sum is σ = 11114311311122. Its totient is φ = 11114311311120.

The previous prime is 11114311311029. The next prime is 11114311311193. The reversal of 11114311311121 is 12111311341111.

It is a m-pointer prime, because the next prime (11114311311193) can be obtained adding 11114311311121 to its product of digits (72).

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 9679167210496 + 1435144100625 = 3111136^2 + 1197975^2 .

It is an emirp because it is prime and its reverse (12111311341111) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 11114311311121 - 2¹³ = 11114311302929 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 11114311311092 and 11114311311101.

It is not a weakly prime, because it can be changed into another prime (11114311411121) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5557155655560 + 5557155655561.

It is an arithmetic number, because the mean of its divisors is an integer number (5557155655561).

Almost surely, 2^{11114311311121} is an apocalyptic number.

It is an amenable number.

11114311311121 is a deficient number, since it is larger than the sum of its proper divisors (1).

11114311311121 is an equidigital number, since it uses as much as digits as its factorization.

11114311311121 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 72, while the sum is 22.

Adding to 11114311311121 its reverse (12111311341111), we get a palindrome (23225622652232).

The spelling of 11114311311121 in words is "eleven trillion, one hundred fourteen billion, three hundred eleven million, three hundred eleven thousand, one hundred twenty-one".