413011300514441 has 2 divisors, whose sum is σ = 413011300514442. Its totient is φ = 413011300514440.

The previous prime is 413011300514419. The next prime is 413011300514443. The reversal of 413011300514441 is 144415003110314.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 391235740902400 + 21775559612041 = 19779680^2 + 4666429^2 .

It is a cyclic number.

It is not a de Polignac number, because 413011300514441 - 2¹⁴ = 413011300498057 is a prime.

Together with 413011300514443, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 413011300514398 and 413011300514407.

It is not a weakly prime, because it can be changed into another prime (413011300514443) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 206505650257220 + 206505650257221.

It is an arithmetic number, because the mean of its divisors is an integer number (206505650257221).

Almost surely, 2^{413011300514441} is an apocalyptic number.

It is an amenable number.

413011300514441 is a deficient number, since it is larger than the sum of its proper divisors (1).

413011300514441 is an equidigital number, since it uses as much as digits as its factorization.

413011300514441 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 11520, while the sum is 32.

Adding to 413011300514441 its reverse (144415003110314), we get a palindrome (557426303624755).

The spelling of 413011300514441 in words is "four hundred thirteen trillion, eleven billion, three hundred million, five hundred fourteen thousand, four hundred forty-one".