145403234341 has 4 divisors (see below), whose sum is σ = 147867696000. Its totient is φ = 142938772684.

The previous prime is 145403234329. The next prime is 145403234363. The reversal of 145403234341 is 143432304541.

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, and also an emirpimes, since its reverse is a distinct semiprime: 143432304541 = 53 ⋅2706269897.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-145403234341 is a prime.

It is a super-2 number, since 2×145403234341² (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 = 145403234297 and 145403234306.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (145403234311) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1232230741 + ... + 1232230858.

It is an arithmetic number, because the mean of its divisors is an integer number (36966924000).

Almost surely, 2^145403234341 is an apocalyptic number.

It is an amenable number.

145403234341 is a deficient number, since it is larger than the sum of its proper divisors (2464461659).

145403234341 is an equidigital number, since it uses as much as digits as its factorization.

145403234341 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 2464461658.

The product of its (nonzero) digits is 69120, while the sum is 34.

Adding to 145403234341 its reverse (143432304541), we get a palindrome (288835538882).

The spelling of 145403234341 in words is "one hundred forty-five billion, four hundred three million, two hundred thirty-four thousand, three hundred forty-one".