131403404131042 has 4 divisors (see below), whose sum is σ = 197105106196566. Its totient is φ = 65701702065520.

The previous prime is 131403404130997. The next prime is 131403404131049. The reversal of 131403404131042 is 240131404304131.

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

It can be written as a sum of positive squares in only one way, i.e., 71311590055161 + 60091814075881 = 8444619^2 + 7751891^2 .

It is a super-3 number, since 3×131403404131042³ (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a junction number, because it is equal to n+sod(n) for n = 131403404130995 and 131403404131013.

It is not an unprimeable number, because it can be changed into a prime (131403404131049) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 32850851032759 + ... + 32850851032762.

Almost surely, 2^{131403404131042} is an apocalyptic number.

131403404131042 is a deficient number, since it is larger than the sum of its proper divisors (65701702065524).

131403404131042 is an equidigital number, since it uses as much as digits as its factorization.

131403404131042 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 65701702065523.

The product of its (nonzero) digits is 13824, while the sum is 31.

Adding to 131403404131042 its reverse (240131404304131), we get a palindrome (371534808435173).

The spelling of 131403404131042 in words is "one hundred thirty-one trillion, four hundred three billion, four hundred four million, one hundred thirty-one thousand, forty-two".