143201234131 has 4 divisors (see below), whose sum is σ = 143272729512. Its totient is φ = 143129738752.

The previous prime is 143201234069. The next prime is 143201234147. The reversal of 143201234131 is 131432102341.

It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 131432102341 = 71 ⋅1851156371.

It is a cyclic number.

It is not a de Polignac number, because 143201234131 - 2¹³ = 143201225939 is a prime.

It is a super-3 number, since 3×143201234131³ (a number of 34 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 143201234096 and 143201234105.

It is not an unprimeable number, because it can be changed into a prime (143201264131) 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, 35744686 + ... + 35748691.

It is an arithmetic number, because the mean of its divisors is an integer number (35818182378).

Almost surely, 2^143201234131 is an apocalyptic number.

143201234131 is a deficient number, since it is larger than the sum of its proper divisors (71495381).

143201234131 is an equidigital number, since it uses as much as digits as its factorization.

143201234131 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 71495380.

The product of its (nonzero) digits is 1728, while the sum is 25.

Adding to 143201234131 its reverse (131432102341), we get a palindrome (274633336472).

The spelling of 143201234131 in words is "one hundred forty-three billion, two hundred one million, two hundred thirty-four thousand, one hundred thirty-one".