1310011101491 has 2 divisors, whose sum is σ = 1310011101492. Its totient is φ = 1310011101490.

The previous prime is 1310011101463. The next prime is 1310011101493. The reversal of 1310011101491 is 1941011100131.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 1310011101491 - 2³⁸ = 1035133194547 is a prime.

It is a super-2 number, since 2×1310011101491² (a number of 25 digits) contains 22 as substring.

Together with 1310011101493, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (1310011101493) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 655005550745 + 655005550746.

It is an arithmetic number, because the mean of its divisors is an integer number (655005550746).

Almost surely, 2^{1310011101491} is an apocalyptic number.

1310011101491 is a deficient number, since it is larger than the sum of its proper divisors (1).

1310011101491 is an equidigital number, since it uses as much as digits as its factorization.

1310011101491 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 108, while the sum is 23.

The spelling of 1310011101491 in words is "one trillion, three hundred ten billion, eleven million, one hundred one thousand, four hundred ninety-one".