201503113131 has 4 divisors (see below), whose sum is σ = 268670817512. Its totient is φ = 134335408752.

The previous prime is 201503113099. The next prime is 201503113139. The reversal of 201503113131 is 131311305102.

It is a semiprime because it is the product of two primes.

It is not a de Polignac number, because 201503113131 - 2⁵ = 201503113099 is a prime.

It is a super-2 number, since 2×201503113131² (a number of 23 digits) contains 22 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 201503113098 and 201503113107.

It is not an unprimeable number, because it can be changed into a prime (201503113139) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 33583852186 + ... + 33583852191.

It is an arithmetic number, because the mean of its divisors is an integer number (67167704378).

Almost surely, 2^201503113131 is an apocalyptic number.

201503113131 is a deficient number, since it is larger than the sum of its proper divisors (67167704381).

201503113131 is an equidigital number, since it uses as much as digits as its factorization.

201503113131 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 67167704380.

The product of its (nonzero) digits is 270, while the sum is 21.

Adding to 201503113131 its reverse (131311305102), we get a palindrome (332814418233).

The spelling of 201503113131 in words is "two hundred one billion, five hundred three million, one hundred thirteen thousand, one hundred thirty-one".