1131314044183 has 2 divisors, whose sum is σ = 1131314044184. Its totient is φ = 1131314044182.

The previous prime is 1131314044169. The next prime is 1131314044321. The reversal of 1131314044183 is 3814404131311.

It is a weak prime.

It is an emirp because it is prime and its reverse (3814404131311) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-1131314044183 is a prime.

It is a super-2 number, since 2×1131314044183² (a number of 25 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1131314044123) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 565657022091 + 565657022092.

It is an arithmetic number, because the mean of its divisors is an integer number (565657022092).

Almost surely, 2^{1131314044183} is an apocalyptic number.

1131314044183 is a deficient number, since it is larger than the sum of its proper divisors (1).

1131314044183 is an equidigital number, since it uses as much as digits as its factorization.

1131314044183 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 13824, while the sum is 34.

Adding to 1131314044183 its reverse (3814404131311), we get a palindrome (4945718175494).

The spelling of 1131314044183 in words is "one trillion, one hundred thirty-one billion, three hundred fourteen million, forty-four thousand, one hundred eighty-three".