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321404212402307 is a prime number
BaseRepresentation
bin100100100010100001011111…
…1111010001101100010000011
31120010222221110011210002121222
41021011002333322031202003
5314111341311333333212
63055323020135305255
7124461455001136226
oct11105027772154203
91503887404702558
10321404212402307
1193455840a01a71
123006a35851b22b
1310a45396c21558
1459520929035bd
1527256db583c72
hex12450bfe8d883

321404212402307 has 2 divisors, whose sum is σ = 321404212402308. Its totient is φ = 321404212402306.

The previous prime is 321404212402267. The next prime is 321404212402309. The reversal of 321404212402307 is 703204212404123.

It is a happy number.

It is a strong prime.

It is an emirp because it is prime and its reverse (703204212404123) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 321404212402307 - 26 = 321404212402243 is a prime.

Together with 321404212402309, 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 (321404212402309) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 160702106201153 + 160702106201154.

It is an arithmetic number, because the mean of its divisors is an integer number (160702106201154).

Almost surely, 2321404212402307 is an apocalyptic number.

321404212402307 is a deficient number, since it is larger than the sum of its proper divisors (1).

321404212402307 is an equidigital number, since it uses as much as digits as its factorization.

321404212402307 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 64512, while the sum is 35.

The spelling of 321404212402307 in words is "three hundred twenty-one trillion, four hundred four billion, two hundred twelve million, four hundred two thousand, three hundred seven".