354099113 has 2 divisors, whose sum is σ = 354099114. Its totient is φ = 354099112.

The previous prime is 354099101. The next prime is 354099121. The reversal of 354099113 is 311990453.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 351825049 + 2274064 = 18757^2 + 1508^2 .

It is an emirp because it is prime and its reverse (311990453) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 354099113 - 2²² = 349904809 is a prime.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (354099913) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 177049556 + 177049557.

It is an arithmetic number, because the mean of its divisors is an integer number (177049557).

Almost surely, 2^354099113 is an apocalyptic number.

It is an amenable number.

354099113 is a deficient number, since it is larger than the sum of its proper divisors (1).

354099113 is an equidigital number, since it uses as much as digits as its factorization.

354099113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 14580, while the sum is 35.

The square root of 354099113 is about 18817.5214361509. The cubic root of 354099113 is about 707.4704090976.

The spelling of 354099113 in words is "three hundred fifty-four million, ninety-nine thousand, one hundred thirteen".