1311413040101 has 2 divisors, whose sum is σ = 1311413040102. Its totient is φ = 1311413040100.

The previous prime is 1311413040083. The next prime is 1311413040103. The reversal of 1311413040101 is 1010403141131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1012066180225 + 299346859876 = 1006015^2 + 547126^2 .

It is a cyclic number.

It is not a de Polignac number, because 1311413040101 - 2²⁶ = 1311345931237 is a prime.

Together with 1311413040103, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1311413040103) 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 as a sum of consecutive naturals, namely, 655706520050 + 655706520051.

It is an arithmetic number, because the mean of its divisors is an integer number (655706520051).

Almost surely, 2^{1311413040101} is an apocalyptic number.

It is an amenable number.

1311413040101 is a deficient number, since it is larger than the sum of its proper divisors (1).

1311413040101 is an equidigital number, since it uses as much as digits as its factorization.

1311413040101 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 144, while the sum is 20.

Adding to 1311413040101 its reverse (1010403141131), we get a palindrome (2321816181232).

The spelling of 1311413040101 in words is "one trillion, three hundred eleven billion, four hundred thirteen million, forty thousand, one hundred one".