101122100113 has 2 divisors, whose sum is σ = 101122100114. Its totient is φ = 101122100112.

The previous prime is 101122100087. The next prime is 101122100117. The reversal of 101122100113 is 311001221101.

It is a happy number.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 98465419264 + 2656680849 = 313792^2 + 51543^2 .

It is an emirp because it is prime and its reverse (311001221101) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 101122100113 - 2⁹ = 101122099601 is a prime.

It is a super-2 number, since 2×101122100113² (a number of 23 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 101122100093 and 101122100102.

It is not a weakly prime, because it can be changed into another prime (101122100117) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50561050056 + 50561050057.

It is an arithmetic number, because the mean of its divisors is an integer number (50561050057).

Almost surely, 2^101122100113 is an apocalyptic number.

It is an amenable number.

101122100113 is a deficient number, since it is larger than the sum of its proper divisors (1).

101122100113 is an equidigital number, since it uses as much as digits as its factorization.

101122100113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 12, while the sum is 13.

Adding to 101122100113 its reverse (311001221101), we get a palindrome (412123321214).

The spelling of 101122100113 in words is "one hundred one billion, one hundred twenty-two million, one hundred thousand, one hundred thirteen".