1311102060881 has 2 divisors, whose sum is σ = 1311102060882. Its totient is φ = 1311102060880.

The previous prime is 1311102060857. The next prime is 1311102060883. The reversal of 1311102060881 is 1880602011131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 743250894400 + 567851166481 = 862120^2 + 753559^2 .

It is an emirp because it is prime and its reverse (1880602011131) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-1311102060881 is a prime.

It is a super-3 number, since 3×1311102060881³ (a number of 37 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

Together with 1311102060883, 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 (1311102060883) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 655551030440 + 655551030441.

It is an arithmetic number, because the mean of its divisors is an integer number (655551030441).

Almost surely, 2^{1311102060881} is an apocalyptic number.

It is an amenable number.

1311102060881 is a deficient number, since it is larger than the sum of its proper divisors (1).

1311102060881 is an equidigital number, since it uses as much as digits as its factorization.

1311102060881 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 2304, while the sum is 32.

The spelling of 1311102060881 in words is "one trillion, three hundred eleven billion, one hundred two million, sixty thousand, eight hundred eighty-one".