3112141031 has 2 divisors, whose sum is σ = 3112141032. Its totient is φ = 3112141030.

The previous prime is 3112141013. The next prime is 3112141033. The reversal of 3112141031 is 1301412113.

Together with previous prime (3112141013) it forms an Ormiston pair, because they use the same digits, order apart.

It is a strong prime.

It is an emirp because it is prime and its reverse (1301412113) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 3112141031 - 2²² = 3107946727 is a prime.

It is a super-2 number, since 2×3112141031² = 19370843593667485922, which contains 22 as substring.

Together with 3112141033, 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 (3112141033) 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, 1556070515 + 1556070516.

It is an arithmetic number, because the mean of its divisors is an integer number (1556070516).

Almost surely, 2^3112141031 is an apocalyptic number.

3112141031 is a deficient number, since it is larger than the sum of its proper divisors (1).

3112141031 is an equidigital number, since it uses as much as digits as its factorization.

3112141031 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 72, while the sum is 17.

The square root of 3112141031 is about 55786.5667611836. The cubic root of 3112141031 is about 1460.0007867325. Note that the first 3 decimals are identical.

Adding to 3112141031 its reverse (1301412113), we get a palindrome (4413553144).

The spelling of 3112141031 in words is "three billion, one hundred twelve million, one hundred forty-one thousand, thirty-one".