1311140114 has 4 divisors (see below), whose sum is σ = 1966710174. Its totient is φ = 655570056.

The previous prime is 1311140111. The next prime is 1311140119. The reversal of 1311140114 is 4110411131.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in only one way, i.e., 1254363889 + 56776225 = 35417^2 + 7535^2 .

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 1311140092 and 1311140101.

It is not an unprimeable number, because it can be changed into a prime (1311140111) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (13) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 327785027 + ... + 327785030.

Almost surely, 2^1311140114 is an apocalyptic number.

1311140114 is a deficient number, since it is larger than the sum of its proper divisors (655570060).

1311140114 is an equidigital number, since it uses as much as digits as its factorization.

1311140114 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 655570059.

The product of its (nonzero) digits is 48, while the sum is 17.

The square root of 1311140114 is about 36209.6687916363. The cubic root of 1311140114 is about 1094.5015179981.

Adding to 1311140114 its reverse (4110411131), we get a palindrome (5421551245).

The spelling of 1311140114 in words is "one billion, three hundred eleven million, one hundred forty thousand, one hundred fourteen".