11451635406313 has 2 divisors, whose sum is σ = 11451635406314. Its totient is φ = 11451635406312.

The previous prime is 11451635406311. The next prime is 11451635406331. The reversal of 11451635406313 is 31360453615411.

Together with next prime (11451635406331) it forms an Ormiston pair, because they use the same digits, order apart.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 6863184373824 + 4588451032489 = 2619768^2 + 2142067^2 .

It is a cyclic number.

It is not a de Polignac number, because 11451635406313 - 2¹ = 11451635406311 is a prime.

It is a super-3 number, since 3×11451635406313³ (a number of 40 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

Together with 11451635406311, it forms a pair of twin primes.

It is a self number, because there is not a number n which added to its sum of digits gives 11451635406313.

It is not a weakly prime, because it can be changed into another prime (11451635406311) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5725817703156 + 5725817703157.

It is an arithmetic number, because the mean of its divisors is an integer number (5725817703157).

Almost surely, 2^{11451635406313} is an apocalyptic number.

It is an amenable number.

11451635406313 is a deficient number, since it is larger than the sum of its proper divisors (1).

11451635406313 is an equidigital number, since it uses as much as digits as its factorization.

11451635406313 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 388800, while the sum is 43.

The spelling of 11451635406313 in words is "eleven trillion, four hundred fifty-one billion, six hundred thirty-five million, four hundred six thousand, three hundred thirteen".