134983943 has 2 divisors, whose sum is σ = 134983944. Its totient is φ = 134983942.

The previous prime is 134983921. The next prime is 134983949. The reversal of 134983943 is 349389431.

It is a strong prime.

It is an emirp because it is prime and its reverse (349389431) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 134983943 - 2⁸ = 134983687 is a prime.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 134983894 and 134983903.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (134983949) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (11) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67491971 + 67491972.

It is an arithmetic number, because the mean of its divisors is an integer number (67491972).

Almost surely, 2^134983943 is an apocalyptic number.

134983943 is a deficient number, since it is larger than the sum of its proper divisors (1).

134983943 is an equidigital number, since it uses as much as digits as its factorization.

134983943 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 279936, while the sum is 44.

The square root of 134983943 is about 11618.2590348124. The cubic root of 134983943 is about 512.9724446160.

The spelling of 134983943 in words is "one hundred thirty-four million, nine hundred eighty-three thousand, nine hundred forty-three".