Base | Representation |
---|---|
bin | 11101100101011010111111… |
… | …011000111100100100010001 |
3 | 122001200212102121012011101211 |
4 | 131211122333120330210101 |
5 | 114023300404231042232 |
6 | 1140421552152500121 |
7 | 36256331522554045 |
oct | 3545327730744421 |
9 | 561625377164354 |
10 | 130114950252817 |
11 | 38505493367572 |
12 | 127151956aa641 |
13 | 577aa3051366a |
14 | 241b8509b4025 |
15 | 10098c606ca47 |
hex | 7656bf63c911 |
130114950252817 has 2 divisors, whose sum is σ = 130114950252818. Its totient is φ = 130114950252816.
The previous prime is 130114950252809. The next prime is 130114950252851. The reversal of 130114950252817 is 718252059411031.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 91500110293761 + 38614839959056 = 9565569^2 + 6214084^2 .
It is a cyclic number.
It is not a de Polignac number, because 130114950252817 - 23 = 130114950252809 is a prime.
It is not a weakly prime, because it can be changed into another prime (130114950252857) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65057475126408 + 65057475126409.
It is an arithmetic number, because the mean of its divisors is an integer number (65057475126409).
Almost surely, 2130114950252817 is an apocalyptic number.
It is an amenable number.
130114950252817 is a deficient number, since it is larger than the sum of its proper divisors (1).
130114950252817 is an equidigital number, since it uses as much as digits as its factorization.
130114950252817 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 604800, while the sum is 49.
The spelling of 130114950252817 in words is "one hundred thirty trillion, one hundred fourteen billion, nine hundred fifty million, two hundred fifty-two thousand, eight hundred seventeen".
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