1201300000037 has 2 divisors, whose sum is σ = 1201300000038. Its totient is φ = 1201300000036.

The previous prime is 1201299999961. The next prime is 1201300000039. The reversal of 1201300000037 is 7300000031021.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1091744957956 + 109555042081 = 1044866^2 + 330991^2 .

It is a cyclic number.

It is not a de Polignac number, because 1201300000037 - 2⁴⁰ = 101788372261 is a prime.

Together with 1201300000039, 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 = 1201299999970 and 1201300000024.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1201300000039) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 600650000018 + 600650000019.

It is an arithmetic number, because the mean of its divisors is an integer number (600650000019).

Almost surely, 2^{1201300000037} is an apocalyptic number.

It is an amenable number.

1201300000037 is a deficient number, since it is larger than the sum of its proper divisors (1).

1201300000037 is an equidigital number, since it uses as much as digits as its factorization.

1201300000037 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 126, while the sum is 17.

Adding to 1201300000037 its reverse (7300000031021), we get a palindrome (8501300031058).

The spelling of 1201300000037 in words is "one trillion, two hundred one billion, three hundred million, thirty-seven", and thus it is an aban number.