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1311320111209 is a prime number
BaseRepresentation
bin10011000101010000110…
…001001111100001101001
311122100202011010200111021
4103011100301033201221
5132441040422024314
62442225004532441
7163511510015143
oct23052061174151
94570664120437
101311320111209
114661450202a9
12192186551121
13968704417b4
144767ab85c93
152419cb47324
hex13150c4f869

1311320111209 has 2 divisors, whose sum is σ = 1311320111210. Its totient is φ = 1311320111208.

The previous prime is 1311320111207. The next prime is 1311320111237. The reversal of 1311320111209 is 9021110231131.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 1140794886400 + 170525224809 = 1068080^2 + 412947^2 .

It is a cyclic number.

It is not a de Polignac number, because 1311320111209 - 21 = 1311320111207 is a prime.

It is a super-2 number, since 2×13113201112092 (a number of 25 digits) contains 22 as substring.

Together with 1311320111207, it forms a pair of twin primes.

It is not a weakly prime, because it can be changed into another prime (1311320111203) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 655660055604 + 655660055605.

It is an arithmetic number, because the mean of its divisors is an integer number (655660055605).

Almost surely, 21311320111209 is an apocalyptic number.

It is an amenable number.

1311320111209 is a deficient number, since it is larger than the sum of its proper divisors (1).

1311320111209 is an equidigital number, since it uses as much as digits as its factorization.

1311320111209 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 324, while the sum is 25.

The spelling of 1311320111209 in words is "one trillion, three hundred eleven billion, three hundred twenty million, one hundred eleven thousand, two hundred nine".