130003543450015 has 4 divisors (see below), whose sum is σ = 156004252140024. Its totient is φ = 104002834760008.

The previous prime is 130003543449911. The next prime is 130003543450037. The reversal of 130003543450015 is 510054345300031.

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

It is a cyclic number.

It is not a de Polignac number, because 130003543450015 - 2¹³ = 130003543441823 is a prime.

It is a Duffinian number.

It is a congruent number.

It is an unprimeable number.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 13000354344997 + ... + 13000354345006.

It is an arithmetic number, because the mean of its divisors is an integer number (39001063035006).

Almost surely, 2^{130003543450015} is an apocalyptic number.

130003543450015 is a deficient number, since it is larger than the sum of its proper divisors (26000708690009).

130003543450015 is an equidigital number, since it uses as much as digits as its factorization.

130003543450015 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 26000708690008.

The product of its (nonzero) digits is 54000, while the sum is 34.

Adding to 130003543450015 its reverse (510054345300031), we get a palindrome (640057888750046).

The spelling of 130003543450015 in words is "one hundred thirty trillion, three billion, five hundred forty-three million, four hundred fifty thousand, fifteen".