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2013143201701 is a prime number
BaseRepresentation
bin11101010010111000101…
…011110111011110100101
321010110021102002200020021
4131102320223313132211
5230440404124423301
64140454125235141
7265305364606252
oct35227053673645
97113242080207
102013143201701
11706850944855
122861b1702ab1
13117ab94c47b8
146d618254429
1537576cb98a1
hex1d4b8af77a5

2013143201701 has 2 divisors, whose sum is σ = 2013143201702. Its totient is φ = 2013143201700.

The previous prime is 2013143201689. The next prime is 2013143201731. The reversal of 2013143201701 is 1071023413102.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 1489683716676 + 523459485025 = 1220526^2 + 723505^2 .

It is a cyclic number.

It is not a de Polignac number, because 2013143201701 - 29 = 2013143201189 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (2013143201731) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1006571600850 + 1006571600851.

It is an arithmetic number, because the mean of its divisors is an integer number (1006571600851).

Almost surely, 22013143201701 is an apocalyptic number.

It is an amenable number.

2013143201701 is a deficient number, since it is larger than the sum of its proper divisors (1).

2013143201701 is an equidigital number, since it uses as much as digits as its factorization.

2013143201701 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1008, while the sum is 25.

Adding to 2013143201701 its reverse (1071023413102), we get a palindrome (3084166614803).

The spelling of 2013143201701 in words is "two trillion, thirteen billion, one hundred forty-three million, two hundred one thousand, seven hundred one".