Base | Representation |
---|---|
bin | 10010111110110101001… |
… | …010001011001110011001 |
3 | 11121200220212212221112011 |
4 | 102332311022023032121 |
5 | 132332420043343301 |
6 | 2435123440145521 |
7 | 163145414515054 |
oct | 22766512131631 |
9 | 4550825787464 |
10 | 1304414434201 |
11 | 463220a3aa1a |
12 | 1909799082a1 |
13 | 9600c842398 |
14 | 471c397da9b |
15 | 23de6760a51 |
hex | 12fb528b399 |
1304414434201 has 2 divisors, whose sum is σ = 1304414434202. Its totient is φ = 1304414434200.
The previous prime is 1304414434133. The next prime is 1304414434207. The reversal of 1304414434201 is 1024344144031.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1100508000601 + 203906433600 = 1049051^2 + 451560^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1304414434201 is a prime.
It is not a weakly prime, because it can be changed into another prime (1304414434207) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 652207217100 + 652207217101.
It is an arithmetic number, because the mean of its divisors is an integer number (652207217101).
Almost surely, 21304414434201 is an apocalyptic number.
It is an amenable number.
1304414434201 is a deficient number, since it is larger than the sum of its proper divisors (1).
1304414434201 is an equidigital number, since it uses as much as digits as its factorization.
1304414434201 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 18432, while the sum is 31.
Adding to 1304414434201 its reverse (1024344144031), we get a palindrome (2328758578232).
The spelling of 1304414434201 in words is "one trillion, three hundred four billion, four hundred fourteen million, four hundred thirty-four thousand, two hundred one".
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