131141342140811 has 2 divisors, whose sum is σ = 131141342140812. Its totient is φ = 131141342140810.

The previous prime is 131141342140801. The next prime is 131141342140813. The reversal of 131141342140811 is 118041243141131.

It is a strong prime.

It is an emirp because it is prime and its reverse (118041243141131) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-131141342140811 is a prime.

Together with 131141342140813, 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 (131141342140813) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65570671070405 + 65570671070406.

It is an arithmetic number, because the mean of its divisors is an integer number (65570671070406).

Almost surely, 2^{131141342140811} is an apocalyptic number.

131141342140811 is a deficient number, since it is larger than the sum of its proper divisors (1).

131141342140811 is an equidigital number, since it uses as much as digits as its factorization.

131141342140811 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 9216, while the sum is 35.

Adding to 131141342140811 its reverse (118041243141131), we get a palindrome (249182585281942).

The spelling of 131141342140811 in words is "one hundred thirty-one trillion, one hundred forty-one billion, three hundred forty-two million, one hundred forty thousand, eight hundred eleven".