Base | Representation |
---|---|
bin | 11101110101001110100101… |
… | …011101101100010000011001 |
3 | 122012112201002011121202211011 |
4 | 131311032211131230100121 |
5 | 114144044414323440001 |
6 | 1143012550023533521 |
7 | 36430653415320352 |
oct | 3565164535542031 |
9 | 565481064552734 |
10 | 131201142015001 |
11 | 38894101296073 |
12 | 1286b7b0bba2a1 |
13 | 58292a36415c6 |
14 | 2458252240d29 |
15 | 1027c99668d51 |
hex | 7753a576c419 |
131201142015001 has 2 divisors, whose sum is σ = 131201142015002. Its totient is φ = 131201142015000.
The previous prime is 131201142014959. The next prime is 131201142015037. The reversal of 131201142015001 is 100510241102131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 105857368267401 + 25343773747600 = 10288701^2 + 5034260^2 .
It is a cyclic number.
It is not a de Polignac number, because 131201142015001 - 227 = 131201007797273 is a prime.
It is not a weakly prime, because it can be changed into another prime (131201142016001) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65600571007500 + 65600571007501.
It is an arithmetic number, because the mean of its divisors is an integer number (65600571007501).
Almost surely, 2131201142015001 is an apocalyptic number.
It is an amenable number.
131201142015001 is a deficient number, since it is larger than the sum of its proper divisors (1).
131201142015001 is an equidigital number, since it uses as much as digits as its factorization.
131201142015001 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 240, while the sum is 22.
Adding to 131201142015001 its reverse (100510241102131), we get a palindrome (231711383117132).
The spelling of 131201142015001 in words is "one hundred thirty-one trillion, two hundred one billion, one hundred forty-two million, fifteen thousand, one".
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