Base | Representation |
---|---|
bin | 10011000101001101100… |
… | …000011001110110011001 |
3 | 11122100121100010022100202 |
4 | 103011031200121312121 |
5 | 132440432413143101 |
6 | 2442215331311545 |
7 | 163510243463315 |
oct | 23051540316631 |
9 | 4570540108322 |
10 | 1311265365401 |
11 | 466117128a32 |
12 | 19217014b5b5 |
13 | 96861cb4224 |
14 | 476737b4b45 |
15 | 24197e3136b |
hex | 1314d819d99 |
1311265365401 has 4 divisors (see below), whose sum is σ = 1311585893904. Its totient is φ = 1310944836900.
The previous prime is 1311265365331. The next prime is 1311265365437. The reversal of 1311265365401 is 1045635621131.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is not a de Polignac number, because 1311265365401 - 210 = 1311265364377 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (1311265365461) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 160258115 + ... + 160266296.
It is an arithmetic number, because the mean of its divisors is an integer number (327896473476).
Almost surely, 21311265365401 is an apocalyptic number.
It is an amenable number.
1311265365401 is a deficient number, since it is larger than the sum of its proper divisors (320528503).
1311265365401 is an equidigital number, since it uses as much as digits as its factorization.
1311265365401 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 320528502.
The product of its (nonzero) digits is 64800, while the sum is 38.
The spelling of 1311265365401 in words is "one trillion, three hundred eleven billion, two hundred sixty-five million, three hundred sixty-five thousand, four hundred one".
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