131115312231133 has 4 divisors (see below), whose sum is σ = 131123835636768. Its totient is φ = 131106788825500.

The previous prime is 131115312231107. The next prime is 131115312231139. The reversal of 131115312231133 is 331132213511131.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 131115312231133 - 2⁴¹ = 128916288975581 is a prime.

It is a super-3 number, since 3×131115312231133³ (a number of 43 digits) contains 333 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 131115312231095 and 131115312231104.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (131115312231139) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (29) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 4261679743 + ... + 4261710508.

It is an arithmetic number, because the mean of its divisors is an integer number (32780958909192).

Almost surely, 2^{131115312231133} is an apocalyptic number.

It is an amenable number.

131115312231133 is a deficient number, since it is larger than the sum of its proper divisors (8523405635).

131115312231133 is an equidigital number, since it uses as much as digits as its factorization.

131115312231133 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 8523405634.

The product of its digits is 4860, while the sum is 31.

Adding to 131115312231133 its reverse (331132213511131), we get a palindrome (462247525742264).

The spelling of 131115312231133 in words is "one hundred thirty-one trillion, one hundred fifteen billion, three hundred twelve million, two hundred thirty-one thousand, one hundred thirty-three".