1011221413003 has 2 divisors, whose sum is σ = 1011221413004. Its totient is φ = 1011221413002.

The previous prime is 1011221412967. The next prime is 1011221413043. The reversal of 1011221413003 is 3003141221101.

It is a weak prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-1011221413003 is a prime.

It is a super-2 number, since 2×1011221413003² (a number of 25 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (1011221413043) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 505610706501 + 505610706502.

It is an arithmetic number, because the mean of its divisors is an integer number (505610706502).

Almost surely, 2^{1011221413003} is an apocalyptic number.

1011221413003 is a deficient number, since it is larger than the sum of its proper divisors (1).

1011221413003 is an equidigital number, since it uses as much as digits as its factorization.

1011221413003 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 144, while the sum is 19.

Adding to 1011221413003 its reverse (3003141221101), we get a palindrome (4014362634104).

The spelling of 1011221413003 in words is "one trillion, eleven billion, two hundred twenty-one million, four hundred thirteen thousand, three".