13825007506001 has 2 divisors, whose sum is σ = 13825007506002. Its totient is φ = 13825007506000.

The previous prime is 13825007505997. The next prime is 13825007506003. The reversal of 13825007506001 is 10060570052831.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 13108020250000 + 716987256001 = 3620500^2 + 846751^2 .

It is an emirp because it is prime and its reverse (10060570052831) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 13825007506001 - 2² = 13825007505997 is a prime.

It is a super-3 number, since 3×13825007506001³ (a number of 40 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

Together with 13825007506003, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (13825007506003) 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, 6912503753000 + 6912503753001.

It is an arithmetic number, because the mean of its divisors is an integer number (6912503753001).

Almost surely, 2^{13825007506001} is an apocalyptic number.

It is an amenable number.

13825007506001 is a deficient number, since it is larger than the sum of its proper divisors (1).

13825007506001 is an equidigital number, since it uses as much as digits as its factorization.

13825007506001 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 50400, while the sum is 38.

Adding to 13825007506001 its reverse (10060570052831), we get a palindrome (23885577558832).

The spelling of 13825007506001 in words is "thirteen trillion, eight hundred twenty-five billion, seven million, five hundred six thousand, one".