1312500113 has 8 divisors (see below), whose sum is σ = 1404034632. Its totient is φ = 1222557696.

The previous prime is 1312500047. The next prime is 1312500179. The reversal of 1312500113 is 3110052131.

It is an interprime number because it is at equal distance from previous prime (1312500047) and next prime (1312500179).

It can be written as a sum of positive squares in 4 ways, for example, as 566773249 + 745726864 = 23807^2 + 27308^2 .

It is a sphenic number, since it is the product of 3 distinct primes.

It is a cyclic number.

It is not a de Polignac number, because 1312500113 - 2²⁸ = 1044064657 is a prime.

It is a Harshad number since it is a multiple of its sum of digits (17).

It is a Duffinian number.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 1312500091 and 1312500100.

It is not an unprimeable number, because it can be changed into a prime (1312500713) by changing a digit.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 396320 + ... + 399617.

It is an arithmetic number, because the mean of its divisors is an integer number (175504329).

Almost surely, 2^1312500113 is an apocalyptic number.

It is an amenable number.

1312500113 is a deficient number, since it is larger than the sum of its proper divisors (91534519).

1312500113 is an equidigital number, since it uses as much as digits as its factorization.

1312500113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 796051.

The product of its (nonzero) digits is 90, while the sum is 17.

The square root of 1312500113 is about 36228.4434250217. The cubic root of 1312500113 is about 1094.8798163933.

Adding to 1312500113 its reverse (3110052131), we get a palindrome (4422552244).

The spelling of 1312500113 in words is "one billion, three hundred twelve million, five hundred thousand, one hundred thirteen".