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134044055255057 is a prime number
BaseRepresentation
bin11110011110100110010000…
…010011000011100000010001
3122120121111010200102211100012
4132132212100103003200101
5120032134211121130212
61153030553333004305
740143235435204463
oct3636462023034021
9576544120384305
10134044055255057
113978a845740aa2
121304a776a83695
1359a43b962cb7c
142515aa44c4133
151076bd8077922
hex79e9904c3811

134044055255057 has 2 divisors, whose sum is σ = 134044055255058. Its totient is φ = 134044055255056.

The previous prime is 134044055255029. The next prime is 134044055255059. The reversal of 134044055255057 is 750552550440431.

It is a happy number.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 111967062939601 + 22076992315456 = 10581449^2 + 4698616^2 .

It is a cyclic number.

It is not a de Polignac number, because 134044055255057 - 214 = 134044055238673 is a prime.

Together with 134044055255059, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 134044055254996 and 134044055255014.

It is not a weakly prime, because it can be changed into another prime (134044055255059) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67022027627528 + 67022027627529.

It is an arithmetic number, because the mean of its divisors is an integer number (67022027627529).

Almost surely, 2134044055255057 is an apocalyptic number.

It is an amenable number.

134044055255057 is a deficient number, since it is larger than the sum of its proper divisors (1).

134044055255057 is an equidigital number, since it uses as much as digits as its factorization.

134044055255057 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 8400000, while the sum is 50.

The spelling of 134044055255057 in words is "one hundred thirty-four trillion, forty-four billion, fifty-five million, two hundred fifty-five thousand, fifty-seven".