Base | Representation |
---|---|
bin | 11101101001100101011011… |
… | …010001101000100111000011 |
3 | 122002201012212020011000000122 |
4 | 131221211123101220213003 |
5 | 114042442304024021144 |
6 | 1141201224024010455 |
7 | 36316110112440422 |
oct | 3551453321504703 |
9 | 562635766130018 |
10 | 130401033423299 |
11 | 38605849901871 |
12 | 12760716300a2b |
13 | 579b9c30338b2 |
14 | 242b62d6cd2b9 |
15 | 101206ba767ee |
hex | 76995b4689c3 |
130401033423299 has 2 divisors, whose sum is σ = 130401033423300. Its totient is φ = 130401033423298.
The previous prime is 130401033423077. The next prime is 130401033423301. The reversal of 130401033423299 is 992324330104031.
It is a strong prime.
It is an emirp because it is prime and its reverse (992324330104031) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 130401033423299 - 232 = 130396738456003 is a prime.
Together with 130401033423301, 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 (130401033420299) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65200516711649 + 65200516711650.
It is an arithmetic number, because the mean of its divisors is an integer number (65200516711650).
Almost surely, 2130401033423299 is an apocalyptic number.
130401033423299 is a deficient number, since it is larger than the sum of its proper divisors (1).
130401033423299 is an equidigital number, since it uses as much as digits as its factorization.
130401033423299 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 419904, while the sum is 44.
The spelling of 130401033423299 in words is "one hundred thirty trillion, four hundred one billion, thirty-three million, four hundred twenty-three thousand, two hundred ninety-nine".
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