Base | Representation |
---|---|
bin | 10011000101010000110… |
… | …001010111100110111101 |
3 | 11122100202011012100220111 |
4 | 103011100301113212331 |
5 | 132441040424104232 |
6 | 2442225005350021 |
7 | 163511510213521 |
oct | 23052061274675 |
9 | 4570664170814 |
10 | 1311320144317 |
11 | 466145043167 |
12 | 192186568311 |
13 | 968704538a1 |
14 | 4767ab93d81 |
15 | 2419cb52047 |
hex | 13150c579bd |
1311320144317 has 2 divisors, whose sum is σ = 1311320144318. Its totient is φ = 1311320144316.
The previous prime is 1311320144279. The next prime is 1311320144333. The reversal of 1311320144317 is 7134410231131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1155584150361 + 155735993956 = 1074981^2 + 394634^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1311320144317 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1311320144917) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 655660072158 + 655660072159.
It is an arithmetic number, because the mean of its divisors is an integer number (655660072159).
Almost surely, 21311320144317 is an apocalyptic number.
It is an amenable number.
1311320144317 is a deficient number, since it is larger than the sum of its proper divisors (1).
1311320144317 is an equidigital number, since it uses as much as digits as its factorization.
1311320144317 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 6048, while the sum is 31.
Adding to 1311320144317 its reverse (7134410231131), we get a palindrome (8445730375448).
The spelling of 1311320144317 in words is "one trillion, three hundred eleven billion, three hundred twenty million, one hundred forty-four thousand, three hundred seventeen".
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