101011000001207 has 2 divisors, whose sum is σ = 101011000001208. Its totient is φ = 101011000001206.

The previous prime is 101011000001123. The next prime is 101011000001209. The reversal of 101011000001207 is 702100000110101.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 101011000001207 - 2¹⁸ = 101010999739063 is a prime.

Together with 101011000001209, 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 = 101011000001191 and 101011000001200.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (101011000001209) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50505500000603 + 50505500000604.

It is an arithmetic number, because the mean of its divisors is an integer number (50505500000604).

Almost surely, 2^{101011000001207} is an apocalyptic number.

101011000001207 is a deficient number, since it is larger than the sum of its proper divisors (1).

101011000001207 is an equidigital number, since it uses as much as digits as its factorization.

101011000001207 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 14, while the sum is 14.

Adding to 101011000001207 its reverse (702100000110101), we get a palindrome (803111000111308).

The spelling of 101011000001207 in words is "one hundred one trillion, eleven billion, one thousand, two hundred seven".