201301401319 has 2 divisors, whose sum is σ = 201301401320. Its totient is φ = 201301401318.

The previous prime is 201301401289. The next prime is 201301401331. The reversal of 201301401319 is 913104103102.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 201301401319 - 2¹³ = 201301393127 is a prime.

It is a super-2 number, since 2×201301401319² (a number of 23 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 201301401293 and 201301401302.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (201301401919) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 100650700659 + 100650700660.

It is an arithmetic number, because the mean of its divisors is an integer number (100650700660).

Almost surely, 2^201301401319 is an apocalyptic number.

201301401319 is a deficient number, since it is larger than the sum of its proper divisors (1).

201301401319 is an equidigital number, since it uses as much as digits as its factorization.

201301401319 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 648, while the sum is 25.

The spelling of 201301401319 in words is "two hundred one billion, three hundred one million, four hundred one thousand, three hundred nineteen".