Base | Representation |
---|---|
bin | 111100101111011100… |
… | …1110011111010010001 |
3 | 110110200200012011011101 |
4 | 1321132321303322101 |
5 | 4114120423411032 |
6 | 135531320144401 |
7 | 12265312115035 |
oct | 1713671637221 |
9 | 413620164141 |
10 | 130441232017 |
11 | 50357747298 |
12 | 21344608101 |
13 | c3ba421675 |
14 | 6455d241c5 |
15 | 35d696dde7 |
hex | 1e5ee73e91 |
130441232017 has 2 divisors, whose sum is σ = 130441232018. Its totient is φ = 130441232016.
The previous prime is 130441231969. The next prime is 130441232027. The reversal of 130441232017 is 710232144031.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 99126744336 + 31314487681 = 314844^2 + 176959^2 .
It is a cyclic number.
It is not a de Polignac number, because 130441232017 - 211 = 130441229969 is a prime.
It is a super-2 number, since 2×1304412320172 (a number of 23 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (130441232027) 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, 65220616008 + 65220616009.
It is an arithmetic number, because the mean of its divisors is an integer number (65220616009).
Almost surely, 2130441232017 is an apocalyptic number.
It is an amenable number.
130441232017 is a deficient number, since it is larger than the sum of its proper divisors (1).
130441232017 is an equidigital number, since it uses as much as digits as its factorization.
130441232017 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 4032, while the sum is 28.
Adding to 130441232017 its reverse (710232144031), we get a palindrome (840673376048).
The spelling of 130441232017 in words is "one hundred thirty billion, four hundred forty-one million, two hundred thirty-two thousand, seventeen".
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