1301412113 has 2 divisors, whose sum is σ = 1301412114. Its totient is φ = 1301412112.

The previous prime is 1301412089. The next prime is 1301412131. The reversal of 1301412113 is 3112141031.

Together with next prime (1301412131) it forms an Ormiston pair, because they use the same digits, order apart.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1279993729 + 21418384 = 35777^2 + 4628^2 .

It is an emirp because it is prime and its reverse (3112141031) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1301412113 - 2⁸ = 1301411857 is a prime.

It is a Sophie Germain prime.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 1301412091 and 1301412100.

It is not a weakly prime, because it can be changed into another prime (1301412173) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 650706056 + 650706057.

It is an arithmetic number, because the mean of its divisors is an integer number (650706057).

Almost surely, 2^1301412113 is an apocalyptic number.

It is an amenable number.

1301412113 is a deficient number, since it is larger than the sum of its proper divisors (1).

1301412113 is an equidigital number, since it uses as much as digits as its factorization.

1301412113 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 1301412113 is about 36075.0899236578. The cubic root of 1301412113 is about 1091.7879118784.

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

The spelling of 1301412113 in words is "one billion, three hundred one million, four hundred twelve thousand, one hundred thirteen".