11004043481 has 2 divisors, whose sum is σ = 11004043482. Its totient is φ = 11004043480.

The previous prime is 11004043429. The next prime is 11004043483. The reversal of 11004043481 is 18434040011.

11004043481 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 7002342400 + 4001701081 = 83680^2 + 63259^2 .

It is a cyclic number.

It is not a de Polignac number, because 11004043481 - 2¹⁰ = 11004042457 is a prime.

It is a super-2 number, since 2×11004043481² (a number of 21 digits) contains 22 as substring.

Together with 11004043483, it forms a pair of twin primes.

It is a Chen prime.

It is a self number, because there is not a number n which added to its sum of digits gives 11004043481.

It is not a weakly prime, because it can be changed into another prime (11004043483) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5502021740 + 5502021741.

It is an arithmetic number, because the mean of its divisors is an integer number (5502021741).

Almost surely, 2^11004043481 is an apocalyptic number.

It is an amenable number.

11004043481 is a deficient number, since it is larger than the sum of its proper divisors (1).

11004043481 is an equidigital number, since it uses as much as digits as its factorization.

11004043481 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1536, while the sum is 26.

Adding to 11004043481 its reverse (18434040011), we get a palindrome (29438083492).

The spelling of 11004043481 in words is "eleven billion, four million, forty-three thousand, four hundred eighty-one".