Base | Representation |
---|---|
bin | 10011000100011010000… |
… | …000101110001011000001 |
3 | 11122021101010110102211122 |
4 | 103010122000232023001 |
5 | 132432200231001032 |
6 | 2441553510122025 |
7 | 163446653210042 |
oct | 23043200561301 |
9 | 4567333412748 |
10 | 1310401422017 |
11 | 465813493536 |
12 | 191b6a950315 |
13 | 9675500351c |
14 | 475d0b628c9 |
15 | 24147178012 |
hex | 1311a02e2c1 |
1310401422017 has 2 divisors, whose sum is σ = 1310401422018. Its totient is φ = 1310401422016.
The previous prime is 1310401422007. The next prime is 1310401422023. The reversal of 1310401422017 is 7102241040131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1120144940161 + 190256481856 = 1058369^2 + 436184^2 .
It is a cyclic number.
It is not a de Polignac number, because 1310401422017 - 236 = 1241681945281 is a prime.
It is not a weakly prime, because it can be changed into another prime (1310401422007) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 655200711008 + 655200711009.
It is an arithmetic number, because the mean of its divisors is an integer number (655200711009).
Almost surely, 21310401422017 is an apocalyptic number.
It is an amenable number.
1310401422017 is a deficient number, since it is larger than the sum of its proper divisors (1).
1310401422017 is an equidigital number, since it uses as much as digits as its factorization.
1310401422017 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1344, while the sum is 26.
Adding to 1310401422017 its reverse (7102241040131), we get a palindrome (8412642462148).
The spelling of 1310401422017 in words is "one trillion, three hundred ten billion, four hundred one million, four hundred twenty-two thousand, seventeen".
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