Base | Representation |
---|---|
bin | 1100001100010111001100… |
… | …0000110001111110001011 |
3 | 1202110122120121100000020111 |
4 | 3003011303000301332023 |
5 | 3224123024230442324 |
6 | 44302512120535151 |
7 | 2552406061101202 |
oct | 303056300617613 |
9 | 52418517300214 |
10 | 13406522515339 |
11 | 42a97464a8438 |
12 | 16063353214b7 |
13 | 7632ca6cb895 |
14 | 344c44570239 |
15 | 183b039cc794 |
hex | c3173031f8b |
13406522515339 has 2 divisors, whose sum is σ = 13406522515340. Its totient is φ = 13406522515338.
The previous prime is 13406522515327. The next prime is 13406522515493. The reversal of 13406522515339 is 93351522560431.
It is a weak prime.
It is an emirp because it is prime and its reverse (93351522560431) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 13406522515339 - 217 = 13406522384267 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 13406522515292 and 13406522515301.
It is not a weakly prime, because it can be changed into another prime (13406522525339) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6703261257669 + 6703261257670.
It is an arithmetic number, because the mean of its divisors is an integer number (6703261257670).
Almost surely, 213406522515339 is an apocalyptic number.
13406522515339 is a deficient number, since it is larger than the sum of its proper divisors (1).
13406522515339 is an equidigital number, since it uses as much as digits as its factorization.
13406522515339 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 2916000, while the sum is 49.
The spelling of 13406522515339 in words is "thirteen trillion, four hundred six billion, five hundred twenty-two million, five hundred fifteen thousand, three hundred thirty-nine".
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