131103142451 has 2 divisors, whose sum is σ = 131103142452. Its totient is φ = 131103142450.

The previous prime is 131103142447. The next prime is 131103142453. The reversal of 131103142451 is 154241301131.

It is a strong prime.

It is an emirp because it is prime and its reverse (154241301131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 131103142451 - 2² = 131103142447 is a prime.

Together with 131103142453, 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 131103142451.

It is not a weakly prime, because it can be changed into another prime (131103142453) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65551571225 + 65551571226.

It is an arithmetic number, because the mean of its divisors is an integer number (65551571226).

Almost surely, 2^131103142451 is an apocalyptic number.

131103142451 is a deficient number, since it is larger than the sum of its proper divisors (1).

131103142451 is an equidigital number, since it uses as much as digits as its factorization.

131103142451 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1440, while the sum is 26.

Adding to 131103142451 its reverse (154241301131), we get a palindrome (285344443582).

The spelling of 131103142451 in words is "one hundred thirty-one billion, one hundred three million, one hundred forty-two thousand, four hundred fifty-one".