1412000113 has 2 divisors, whose sum is σ = 1412000114. Its totient is φ = 1412000112.

The previous prime is 1412000089. The next prime is 1412000131. The reversal of 1412000113 is 3110002141.

Together with next prime (1412000131) 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., 1341976689 + 70023424 = 36633^2 + 8368^2 .

It is a cyclic number.

It is not a de Polignac number, because 1412000113 - 2⁵ = 1412000081 is a prime.

It is a super-2 number, since 2×1412000113² = 3987488638224025538, which contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1412000093 and 1412000102.

It is not a weakly prime, because it can be changed into another prime (1412000173) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 706000056 + 706000057.

It is an arithmetic number, because the mean of its divisors is an integer number (706000057).

Almost surely, 2^1412000113 is an apocalyptic number.

It is an amenable number.

1412000113 is a deficient number, since it is larger than the sum of its proper divisors (1).

1412000113 is an equidigital number, since it uses as much as digits as its factorization.

1412000113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 24, while the sum is 13.

The square root of 1412000113 is about 37576.5899597076. The cubic root of 1412000113 is about 1121.8761372138.

Adding to 1412000113 its reverse (3110002141), we get a palindrome (4522002254).

The spelling of 1412000113 in words is "one billion, four hundred twelve million, one hundred thirteen", and thus it is an aban number.