110069849407993 has 2 divisors, whose sum is σ = 110069849407994. Its totient is φ = 110069849407992.

The previous prime is 110069849407937. The next prime is 110069849408063. The reversal of 110069849407993 is 399704948960011.

It is an a-pointer prime, because the next prime (110069849408063) can be obtained adding 110069849407993 to its sum of digits (70).

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 71324054857449 + 38745794550544 = 8445357^2 + 6224612^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-110069849407993 is a prime.

It is a super-4 number, since 4×110069849407993⁴ (a number of 57 digits) contains 4444 as substring.

It is not a weakly prime, because it can be changed into another prime (110069849407493) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 55034924703996 + 55034924703997.

It is an arithmetic number, because the mean of its divisors is an integer number (55034924703997).

Almost surely, 2^{110069849407993} is an apocalyptic number.

It is an amenable number.

110069849407993 is a deficient number, since it is larger than the sum of its proper divisors (1).

110069849407993 is an equidigital number, since it uses as much as digits as its factorization.

110069849407993 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 105815808, while the sum is 70.

The spelling of 110069849407993 in words is "one hundred ten trillion, sixty-nine billion, eight hundred forty-nine million, four hundred seven thousand, nine hundred ninety-three".