Base | Representation |
---|---|
bin | 1011111011010110101100… |
… | …1111100101110110100001 |
3 | 1201102201101111212001011002 |
4 | 2332311223033211312201 |
5 | 3204331124101412001 |
6 | 43520350403401345 |
7 | 2522323301115245 |
oct | 276655317456641 |
9 | 51381344761132 |
10 | 13114334404001 |
11 | 41a6837976776 |
12 | 1579790083255 |
13 | 7418a415c1cc |
14 | 334a46b2a425 |
15 | 17b20205c46b |
hex | bed6b3e5da1 |
13114334404001 has 2 divisors, whose sum is σ = 13114334404002. Its totient is φ = 13114334404000.
The previous prime is 13114334403983. The next prime is 13114334404007. The reversal of 13114334404001 is 10040443341131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 11106062795776 + 2008271608225 = 3332576^2 + 1417135^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13114334404001 is a prime.
It is not a weakly prime, because it can be changed into another prime (13114334404007) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6557167202000 + 6557167202001.
It is an arithmetic number, because the mean of its divisors is an integer number (6557167202001).
Almost surely, 213114334404001 is an apocalyptic number.
It is an amenable number.
13114334404001 is a deficient number, since it is larger than the sum of its proper divisors (1).
13114334404001 is an equidigital number, since it uses as much as digits as its factorization.
13114334404001 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 6912, while the sum is 29.
Adding to 13114334404001 its reverse (10040443341131), we get a palindrome (23154777745132).
The spelling of 13114334404001 in words is "thirteen trillion, one hundred fourteen billion, three hundred thirty-four million, four hundred four thousand, one".
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